{"config":{"lang":["en"],"separator":"[\\s\\-]+","pipeline":["stopWordFilter"]},"docs":[{"location":"","title":"Home","text":"Hello Algo

Data Structures and Algorithms Crash Course with Animated Illustrations and Off-the-Shelf Code

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The English edition is brewing...

Feel free to engage in Chinese-to-English translation and pull request review! For guidelines, please see #914.

Endorsements

Quote

\"An easy-to-understand book on data structures and algorithms, which guides readers to learn by minds-on and hands-on. Strongly recommended for algorithm beginners!\"

\u2014\u2014 Junhui Deng, Professor of Computer Science, Tsinghua University

Quote

\"If I had 'Hello Algo' when I was learning data structures and algorithms, it would have been 10 times easier!\"

\u2014\u2014 Mu Li, Senior Principal Scientist, Amazon

Animated illustrations

Easy to understandSmooth learning curve

\"A picture is worth a thousand words.\"

Off-the-Shelf Code

Multi programming languagesRun with one click

\"Talk is cheap. Show me the code.\"

Learning Together

Discussion and questions welcomeReaders progress together

\"Chase the wind and moon, never stopping\"

\"Beyond the plains, there are spring mountains\"

Preface

Two years ago, I shared the \"Sword Offer\" series of problem solutions on LeetCode, which received much love and support from many students. During my interactions with readers, the most common question I encountered was \"How to get started with algorithms.\" Gradually, I developed a deep interest in this question.

Blindly solving problems seems to be the most popular method, being simple, direct, and effective. However, problem-solving is like playing a \"Minesweeper\" game, where students with strong self-learning abilities can successfully clear the mines one by one, but those with insufficient foundations may end up bruised from explosions, retreating step by step in frustration. Thoroughly reading textbooks is also common, but for students aiming for job applications, the energy consumed by graduation, resume submissions, and preparing for written tests and interviews makes tackling thick books a daunting challenge.

If you are facing similar troubles, then you are lucky to have found this book. This book is my answer to this question, not necessarily the best solution, but at least an active attempt. Although this book won't directly land you an Offer, it will guide you through the \"knowledge map\" of data structures and algorithms, help you understand the shape, size, and distribution of different \"mines,\" and equip you with various \"demining methods.\" With these skills, I believe you can more comfortably solve problems and read literature, gradually building a complete knowledge system.

I deeply agree with Professor Feynman's saying: \"Knowledge isn't free. You have to pay attention.\" In this sense, this book is not entirely \"free.\" To not disappoint the precious \"attention\" you pay to this book, I will do my utmost, investing the greatest \"attention\" to complete the creation of this book.

Author

Yudong Jin(Krahets), Senior Algorithm Engineer in a top tech company, Master's degree from Shanghai Jiao Tong University. The highest-read blogger across the entire LeetCode, his published \"Illustration of Algorithm Data Structures\" has been subscribed to by over 300k.

Contribution

This book is continuously improved with the joint efforts of many contributors from the open-source community. Thanks to each writer who invested their time and energy, listed in the order generated by GitHub:

The code review work for this book was completed by Gonglja, gvenusleo, hpstory, justin\u2010tse, krahets, night-cruise, nuomi1, Reanon, and sjinzh (listed in alphabetical order). Thanks to them for their time and effort, ensuring the standardization and uniformity of the code in various languages.

GongljaC, C++ gvenusleoDart hpstoryC# justin-tseJS, TS krahetsJava, Python night-cruiseRust nuomi1Swift ReanonGo, C sjinzhRust, Zig"},{"location":"chapter_computational_complexity/","title":"Chapter 2. \u00a0 Complexity Analysis","text":"

Abstract

Complexity analysis is like a space-time navigator in the vast universe of algorithms.

It guides us in exploring deeper within the the dimensions of time and space, seeking more elegant solutions.

"},{"location":"chapter_computational_complexity/#_1","title":"\u672c\u7ae0\u5185\u5bb9","text":""},{"location":"chapter_computational_complexity/iteration_and_recursion/","title":"2.2 \u00a0 Iteration and Recursion","text":"

In algorithms, repeatedly performing a task is common and closely related to complexity analysis. Therefore, before introducing time complexity and space complexity, let's first understand how to implement task repetition in programs, focusing on two basic programming control structures: iteration and recursion.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#221-iteration","title":"2.2.1 \u00a0 Iteration","text":"

\"Iteration\" is a control structure for repeatedly performing a task. In iteration, a program repeats a block of code as long as a certain condition is met, until this condition is no longer satisfied.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#1-for-loop","title":"1. \u00a0 for Loop","text":"

The for loop is one of the most common forms of iteration, suitable for use when the number of iterations is known in advance.

The following function implements the sum \\(1 + 2 + \\dots + n\\) using a for loop, with the sum result recorded in the variable res. Note that in Python, range(a, b) corresponds to a \"left-closed, right-open\" interval, covering \\(a, a + 1, \\dots, b-1\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig iteration.py
def for_loop(n: int) -> int:\n    \"\"\"for \u5faa\u73af\"\"\"\n    res = 0\n    # \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for i in range(1, n + 1):\n        res += i\n    return res\n
iteration.cpp
/* for \u5faa\u73af */\nint forLoop(int n) {\n    int res = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; ++i) {\n        res += i;\n    }\n    return res;\n}\n
iteration.java
/* for \u5faa\u73af */\nint forLoop(int n) {\n    int res = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; i++) {\n        res += i;\n    }\n    return res;\n}\n
iteration.cs
/* for \u5faa\u73af */\nint ForLoop(int n) {\n    int res = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; i++) {\n        res += i;\n    }\n    return res;\n}\n
iteration.go
/* for \u5faa\u73af */\nfunc forLoop(n int) int {\n    res := 0\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for i := 1; i <= n; i++ {\n        res += i\n    }\n    return res\n}\n
iteration.swift
/* for \u5faa\u73af */\nfunc forLoop(n: Int) -> Int {\n    var res = 0\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for i in 1 ... n {\n        res += i\n    }\n    return res\n}\n
iteration.js
/* for \u5faa\u73af */\nfunction forLoop(n) {\n    let res = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for (let i = 1; i <= n; i++) {\n        res += i;\n    }\n    return res;\n}\n
iteration.ts
/* for \u5faa\u73af */\nfunction forLoop(n: number): number {\n    let res = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for (let i = 1; i <= n; i++) {\n        res += i;\n    }\n    return res;\n}\n
iteration.dart
/* for \u5faa\u73af */\nint forLoop(int n) {\n  int res = 0;\n  // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n  for (int i = 1; i <= n; i++) {\n    res += i;\n  }\n  return res;\n}\n
iteration.rs
/* for \u5faa\u73af */\nfn for_loop(n: i32) -> i32 {\n    let mut res = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for i in 1..=n {\n        res += i;\n    }\n    res\n} \n
iteration.c
/* for \u5faa\u73af */\nint forLoop(int n) {\n    int res = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; i++) {\n        res += i;\n    }\n    return res;\n}\n
iteration.zig
// for \u5faa\u73af\nfn forLoop(n: usize) i32 {\n    var res: i32 = 0;\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for (1..n+1) |i| {\n        res = res + @as(i32, @intCast(i));\n    }\n    return res;\n} \n

The flowchart below represents this sum function.

Figure 2-1 \u00a0 Flowchart of the Sum Function

The number of operations in this sum function is proportional to the input data size \\(n\\), or in other words, it has a \"linear relationship\". This is actually what time complexity describes. This topic will be detailed in the next section.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#2-while-loop","title":"2. \u00a0 while Loop","text":"

Similar to the for loop, the while loop is another method to implement iteration. In a while loop, the program checks the condition in each round; if the condition is true, it continues, otherwise, the loop ends.

Below we use a while loop to implement the sum \\(1 + 2 + \\dots + n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig iteration.py
def while_loop(n: int) -> int:\n    \"\"\"while \u5faa\u73af\"\"\"\n    res = 0\n    i = 1  # \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    # \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while i <= n:\n        res += i\n        i += 1  # \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    return res\n
iteration.cpp
/* while \u5faa\u73af */\nint whileLoop(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while (i <= n) {\n        res += i;\n        i++; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    return res;\n}\n
iteration.java
/* while \u5faa\u73af */\nint whileLoop(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while (i <= n) {\n        res += i;\n        i++; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    return res;\n}\n
iteration.cs
/* while \u5faa\u73af */\nint WhileLoop(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while (i <= n) {\n        res += i;\n        i += 1; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    return res;\n}\n
iteration.go
/* while \u5faa\u73af */\nfunc whileLoop(n int) int {\n    res := 0\n    // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    i := 1\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    for i <= n {\n        res += i\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i++\n    }\n    return res\n}\n
iteration.swift
/* while \u5faa\u73af */\nfunc whileLoop(n: Int) -> Int {\n    var res = 0\n    var i = 1 // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while i <= n {\n        res += i\n        i += 1 // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    return res\n}\n
iteration.js
/* while \u5faa\u73af */\nfunction whileLoop(n) {\n    let res = 0;\n    let i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while (i <= n) {\n        res += i;\n        i++; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    return res;\n}\n
iteration.ts
/* while \u5faa\u73af */\nfunction whileLoop(n: number): number {\n    let res = 0;\n    let i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while (i <= n) {\n        res += i;\n        i++; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    return res;\n}\n
iteration.dart
/* while \u5faa\u73af */\nint whileLoop(int n) {\n  int res = 0;\n  int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n  // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n  while (i <= n) {\n    res += i;\n    i++; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n  }\n  return res;\n}\n
iteration.rs
/* while \u5faa\u73af */\nfn while_loop(n: i32) -> i32 {\n    let mut res = 0;\n    let mut i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while i <= n {\n        res += i;\n        i += 1; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    res\n}\n
iteration.c
/* while \u5faa\u73af */\nint whileLoop(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while (i <= n) {\n        res += i;\n        i++; // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    }\n    return res;\n}\n
iteration.zig
// while \u5faa\u73af\nfn whileLoop(n: i32) i32 {\n    var res: i32 = 0;\n    var i: i32 = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, ..., n-1, n\n    while (i <= n) {\n        res += @intCast(i);\n        i += 1;\n    }\n    return res;\n}\n

The while loop is more flexible than the for loop. In a while loop, we can freely design the initialization and update steps of the condition variable.

For example, in the following code, the condition variable \\(i\\) is updated twice in each round, which would be inconvenient to implement with a for loop:

PythonC++JavaC#GoSwiftJSTSDartRustCZig iteration.py
def while_loop_ii(n: int) -> int:\n    \"\"\"while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09\"\"\"\n    res = 0\n    i = 1  # \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    # \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while i <= n:\n        res += i\n        # \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i += 1\n        i *= 2\n    return res\n
iteration.cpp
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nint whileLoopII(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while (i <= n) {\n        res += i;\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i++;\n        i *= 2;\n    }\n    return res;\n}\n
iteration.java
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nint whileLoopII(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while (i <= n) {\n        res += i;\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i++;\n        i *= 2;\n    }\n    return res;\n}\n
iteration.cs
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nint WhileLoopII(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 2, 4, 5...\n    while (i <= n) {\n        res += i;\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i += 1; \n        i *= 2;\n    }\n    return res;\n}\n
iteration.go
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nfunc whileLoopII(n int) int {\n    res := 0\n    // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    i := 1\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    for i <= n {\n        res += i\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i++\n        i *= 2\n    }\n    return res\n}\n
iteration.swift
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nfunc whileLoopII(n: Int) -> Int {\n    var res = 0\n    var i = 1 // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while i <= n {\n        res += i\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i += 1\n        i *= 2\n    }\n    return res\n}\n
iteration.js
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nfunction whileLoopII(n) {\n    let res = 0;\n    let i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while (i <= n) {\n        res += i;\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i++;\n        i *= 2;\n    }\n    return res;\n}\n
iteration.ts
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nfunction whileLoopII(n: number): number {\n    let res = 0;\n    let i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while (i <= n) {\n        res += i;\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i++;\n        i *= 2;\n    }\n    return res;\n}\n
iteration.dart
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nint whileLoopII(int n) {\n  int res = 0;\n  int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n  // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n  while (i <= n) {\n    res += i;\n    // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n    i++;\n    i *= 2;\n  }\n  return res;\n}\n
iteration.rs
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nfn while_loop_ii(n: i32) -> i32 {\n    let mut res = 0;\n    let mut i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while i <= n {\n        res += i;\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i += 1;\n        i *= 2;\n    }\n    res\n}\n
iteration.c
/* while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09 */\nint whileLoopII(int n) {\n    int res = 0;\n    int i = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while (i <= n) {\n        res += i;\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i++;\n        i *= 2;\n    }\n    return res;\n}\n
iteration.zig
//  while \u5faa\u73af\uff08\u4e24\u6b21\u66f4\u65b0\uff09\nfn whileLoopII(n: i32) i32 {\n    var res: i32 = 0;\n    var i: i32 = 1; // \u521d\u59cb\u5316\u6761\u4ef6\u53d8\u91cf\n    // \u5faa\u73af\u6c42\u548c 1, 4, 10, ...\n    while (i <= n) {\n        res += @intCast(i);\n        // \u66f4\u65b0\u6761\u4ef6\u53d8\u91cf\n        i += 1;\n        i *= 2;\n    }\n    return res;\n}\n

Overall, for loops are more concise, while while loops are more flexible. Both can implement iterative structures. Which one to use should be determined based on the specific requirements of the problem.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#3-nested-loops","title":"3. \u00a0 Nested Loops","text":"

We can nest one loop structure within another. Below is an example using for loops:

PythonC++JavaC#GoSwiftJSTSDartRustCZig iteration.py
def nested_for_loop(n: int) -> str:\n    \"\"\"\u53cc\u5c42 for \u5faa\u73af\"\"\"\n    res = \"\"\n    # \u5faa\u73af i = 1, 2, ..., n-1, n\n    for i in range(1, n + 1):\n        # \u5faa\u73af j = 1, 2, ..., n-1, n\n        for j in range(1, n + 1):\n            res += f\"({i}, {j}), \"\n    return res\n
iteration.cpp
/* \u53cc\u5c42 for \u5faa\u73af */\nstring nestedForLoop(int n) {\n    ostringstream res;\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; ++i) {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for (int j = 1; j <= n; ++j) {\n            res << \"(\" << i << \", \" << j << \"), \";\n        }\n    }\n    return res.str();\n}\n
iteration.java
/* \u53cc\u5c42 for \u5faa\u73af */\nString nestedForLoop(int n) {\n    StringBuilder res = new StringBuilder();\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; i++) {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for (int j = 1; j <= n; j++) {\n            res.append(\"(\" + i + \", \" + j + \"), \");\n        }\n    }\n    return res.toString();\n}\n
iteration.cs
/* \u53cc\u5c42 for \u5faa\u73af */\nstring NestedForLoop(int n) {\n    StringBuilder res = new();\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; i++) {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for (int j = 1; j <= n; j++) {\n            res.Append($\"({i}, {j}), \");\n        }\n    }\n    return res.ToString();\n}\n
iteration.go
/* \u53cc\u5c42 for \u5faa\u73af */\nfunc nestedForLoop(n int) string {\n    res := \"\"\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for i := 1; i <= n; i++ {\n        for j := 1; j <= n; j++ {\n            // \u5faa\u73af j = 1, 2, ..., n-1, n\n            res += fmt.Sprintf(\"(%d, %d), \", i, j)\n        }\n    }\n    return res\n}\n
iteration.swift
/* \u53cc\u5c42 for \u5faa\u73af */\nfunc nestedForLoop(n: Int) -> String {\n    var res = \"\"\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for i in 1 ... n {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for j in 1 ... n {\n            res.append(\"(\\(i), \\(j)), \")\n        }\n    }\n    return res\n}\n
iteration.js
/* \u53cc\u5c42 for \u5faa\u73af */\nfunction nestedForLoop(n) {\n    let res = '';\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for (let i = 1; i <= n; i++) {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for (let j = 1; j <= n; j++) {\n            res += `(${i}, ${j}), `;\n        }\n    }\n    return res;\n}\n
iteration.ts
/* \u53cc\u5c42 for \u5faa\u73af */\nfunction nestedForLoop(n: number): string {\n    let res = '';\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for (let i = 1; i <= n; i++) {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for (let j = 1; j <= n; j++) {\n            res += `(${i}, ${j}), `;\n        }\n    }\n    return res;\n}\n
iteration.dart
/* \u53cc\u5c42 for \u5faa\u73af */\nString nestedForLoop(int n) {\n  String res = \"\";\n  // \u5faa\u73af i = 1, 2, ..., n-1, n\n  for (int i = 1; i <= n; i++) {\n    // \u5faa\u73af j = 1, 2, ..., n-1, n\n    for (int j = 1; j <= n; j++) {\n      res += \"($i, $j), \";\n    }\n  }\n  return res;\n}\n
iteration.rs
/* \u53cc\u5c42 for \u5faa\u73af */\nfn nested_for_loop(n: i32) -> String {\n    let mut res = vec![];\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for i in 1..=n {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for j in 1..=n {\n            res.push(format!(\"({}, {}), \", i, j));\n        }\n    }\n    res.join(\"\")\n}\n
iteration.c
/* \u53cc\u5c42 for \u5faa\u73af */\nchar *nestedForLoop(int n) {\n    // n * n \u4e3a\u5bf9\u5e94\u70b9\u6570\u91cf\uff0c\"(i, j), \" \u5bf9\u5e94\u5b57\u7b26\u4e32\u957f\u6700\u5927\u4e3a 6+10*2\uff0c\u52a0\u4e0a\u6700\u540e\u4e00\u4e2a\u7a7a\u5b57\u7b26 \\0 \u7684\u989d\u5916\u7a7a\u95f4\n    int size = n * n * 26 + 1;\n    char *res = malloc(size * sizeof(char));\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for (int i = 1; i <= n; i++) {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for (int j = 1; j <= n; j++) {\n            char tmp[26];\n            snprintf(tmp, sizeof(tmp), \"(%d, %d), \", i, j);\n            strncat(res, tmp, size - strlen(res) - 1);\n        }\n    }\n    return res;\n}\n
iteration.zig
// \u53cc\u5c42 for \u5faa\u73af\nfn nestedForLoop(allocator: Allocator, n: usize) ![]const u8 {\n    var res = std.ArrayList(u8).init(allocator);\n    defer res.deinit();\n    var buffer: [20]u8 = undefined;\n    // \u5faa\u73af i = 1, 2, ..., n-1, n\n    for (1..n+1) |i| {\n        // \u5faa\u73af j = 1, 2, ..., n-1, n\n        for (1..n+1) |j| {\n            var _str = try std.fmt.bufPrint(&buffer, \"({d}, {d}), \", .{i, j});\n            try res.appendSlice(_str);\n        }\n    }\n    return res.toOwnedSlice();\n}\n

The flowchart below represents this nested loop.

Figure 2-2 \u00a0 Flowchart of the Nested Loop

In this case, the number of operations in the function is proportional to \\(n^2\\), or the algorithm's running time and the input data size \\(n\\) have a \"quadratic relationship\".

We can continue adding nested loops, each nesting is a \"dimensional escalation,\" which will increase the time complexity to \"cubic,\" \"quartic,\" and so on.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#222-recursion","title":"2.2.2 \u00a0 Recursion","text":"

\"Recursion\" is an algorithmic strategy that solves problems by having a function call itself. It mainly consists of two phases.

  1. Recursion: The program continuously calls itself, usually with smaller or more simplified parameters, until reaching a \"termination condition.\"
  2. Return: Upon triggering the \"termination condition,\" the program begins to return from the deepest recursive function, aggregating the results of each layer.

From an implementation perspective, recursive code mainly includes three elements.

  1. Termination Condition: Determines when to switch from \"recursion\" to \"return.\"
  2. Recursive Call: Corresponds to \"recursion,\" where the function calls itself, usually with smaller or more simplified parameters.
  3. Return Result: Corresponds to \"return,\" where the result of the current recursion level is returned to the previous layer.

Observe the following code, where calling the function recur(n) completes the computation of \\(1 + 2 + \\dots + n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig recursion.py
def recur(n: int) -> int:\n    \"\"\"\u9012\u5f52\"\"\"\n    # \u7ec8\u6b62\u6761\u4ef6\n    if n == 1:\n        return 1\n    # \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    res = recur(n - 1)\n    # \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res\n
recursion.cpp
/* \u9012\u5f52 */\nint recur(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 1)\n        return 1;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    int res = recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res;\n}\n
recursion.java
/* \u9012\u5f52 */\nint recur(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 1)\n        return 1;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    int res = recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res;\n}\n
recursion.cs
/* \u9012\u5f52 */\nint Recur(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 1)\n        return 1;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    int res = Recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res;\n}\n
recursion.go
/* \u9012\u5f52 */\nfunc recur(n int) int {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if n == 1 {\n        return 1\n    }\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    res := recur(n - 1)\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res\n}\n
recursion.swift
/* \u9012\u5f52 */\nfunc recur(n: Int) -> Int {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if n == 1 {\n        return 1\n    }\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    let res = recur(n: n - 1)\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res\n}\n
recursion.js
/* \u9012\u5f52 */\nfunction recur(n) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n === 1) return 1;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    const res = recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res;\n}\n
recursion.ts
/* \u9012\u5f52 */\nfunction recur(n: number): number {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n === 1) return 1;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    const res = recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res;\n}\n
recursion.dart
/* \u9012\u5f52 */\nint recur(int n) {\n  // \u7ec8\u6b62\u6761\u4ef6\n  if (n == 1) return 1;\n  // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n  int res = recur(n - 1);\n  // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n  return n + res;\n}\n
recursion.rs
/* \u9012\u5f52 */\nfn recur(n: i32) -> i32 {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if n == 1 {\n        return 1;\n    }\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    let res = recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    n + res\n}\n
recursion.c
/* \u9012\u5f52 */\nint recur(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 1)\n        return 1;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    int res = recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res;\n}\n
recursion.zig
// \u9012\u5f52\u51fd\u6570\nfn recur(n: i32) i32 {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 1) {\n        return 1;\n    }\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    var res: i32 = recur(n - 1);\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    return n + res;\n}\n

The Figure 2-3 shows the recursive process of this function.

Figure 2-3 \u00a0 Recursive Process of the Sum Function

Although iteration and recursion can achieve the same results from a computational standpoint, they represent two entirely different paradigms of thinking and solving problems.

Taking the sum function as an example, let's define the problem as \\(f(n) = 1 + 2 + \\dots + n\\).

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#1-call-stack","title":"1. \u00a0 Call Stack","text":"

Each time a recursive function calls itself, the system allocates memory for the newly initiated function to store local variables, call addresses, and other information. This leads to two main consequences.

As shown in the Figure 2-4 , there are \\(n\\) unreturned recursive functions before triggering the termination condition, indicating a recursion depth of \\(n\\).

Figure 2-4 \u00a0 Recursion Call Depth

In practice, the depth of recursion allowed by programming languages is usually limited, and excessively deep recursion can lead to stack overflow errors.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#2-tail-recursion","title":"2. \u00a0 Tail Recursion","text":"

Interestingly, if a function makes its recursive call as the last step before returning, it can be optimized by compilers or interpreters to be as space-efficient as iteration. This scenario is known as \"tail recursion\".

For example, in calculating \\(1 + 2 + \\dots + n\\), we can make the result variable res a parameter of the function, thereby achieving tail recursion:

PythonC++JavaC#GoSwiftJSTSDartRustCZig recursion.py
def tail_recur(n, res):\n    \"\"\"\u5c3e\u9012\u5f52\"\"\"\n    # \u7ec8\u6b62\u6761\u4ef6\n    if n == 0:\n        return res\n    # \u5c3e\u9012\u5f52\u8c03\u7528\n    return tail_recur(n - 1, res + n)\n
recursion.cpp
/* \u5c3e\u9012\u5f52 */\nint tailRecur(int n, int res) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 0)\n        return res;\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n - 1, res + n);\n}\n
recursion.java
/* \u5c3e\u9012\u5f52 */\nint tailRecur(int n, int res) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 0)\n        return res;\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n - 1, res + n);\n}\n
recursion.cs
/* \u5c3e\u9012\u5f52 */\nint TailRecur(int n, int res) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 0)\n        return res;\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return TailRecur(n - 1, res + n);\n}\n
recursion.go
/* \u5c3e\u9012\u5f52 */\nfunc tailRecur(n int, res int) int {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if n == 0 {\n        return res\n    }\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n-1, res+n)\n}\n
recursion.swift
/* \u5c3e\u9012\u5f52 */\nfunc tailRecur(n: Int, res: Int) -> Int {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if n == 0 {\n        return res\n    }\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n: n - 1, res: res + n)\n}\n
recursion.js
/* \u5c3e\u9012\u5f52 */\nfunction tailRecur(n, res) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n === 0) return res;\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n - 1, res + n);\n}\n
recursion.ts
/* \u5c3e\u9012\u5f52 */\nfunction tailRecur(n: number, res: number): number {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n === 0) return res;\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n - 1, res + n);\n}\n
recursion.dart
/* \u5c3e\u9012\u5f52 */\nint tailRecur(int n, int res) {\n  // \u7ec8\u6b62\u6761\u4ef6\n  if (n == 0) return res;\n  // \u5c3e\u9012\u5f52\u8c03\u7528\n  return tailRecur(n - 1, res + n);\n}\n
recursion.rs
/* \u5c3e\u9012\u5f52 */\nfn tail_recur(n: i32, res: i32) -> i32 {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if n == 0 {\n        return res;\n    }\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    tail_recur(n - 1, res + n)\n}\n
recursion.c
/* \u5c3e\u9012\u5f52 */\nint tailRecur(int n, int res) {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 0)\n        return res;\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n - 1, res + n);\n}\n
recursion.zig
// \u5c3e\u9012\u5f52\u51fd\u6570\nfn tailRecur(n: i32, res: i32) i32 {\n    // \u7ec8\u6b62\u6761\u4ef6\n    if (n == 0) {\n        return res;\n    }\n    // \u5c3e\u9012\u5f52\u8c03\u7528\n    return tailRecur(n - 1, res + n);\n}\n

The execution process of tail recursion is shown in the following figure. Comparing regular recursion and tail recursion, the point of the summation operation is different.

Figure 2-5 \u00a0 Tail Recursion Process

Tip

Note that many compilers or interpreters do not support tail recursion optimization. For example, Python does not support tail recursion optimization by default, so even if the function is in the form of tail recursion, it may still encounter stack overflow issues.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#3-recursion-tree","title":"3. \u00a0 Recursion Tree","text":"

When dealing with algorithms related to \"divide and conquer\", recursion often offers a more intuitive approach and more readable code than iteration. Take the \"Fibonacci sequence\" as an example.

Question

Given a Fibonacci sequence \\(0, 1, 1, 2, 3, 5, 8, 13, \\dots\\), find the \\(n\\)th number in the sequence.

Let the \\(n\\)th number of the Fibonacci sequence be \\(f(n)\\), it's easy to deduce two conclusions:

Using the recursive relation, and considering the first two numbers as termination conditions, we can write the recursive code. Calling fib(n) will yield the \\(n\\)th number of the Fibonacci sequence:

PythonC++JavaC#GoSwiftJSTSDartRustCZig recursion.py
def fib(n: int) -> int:\n    \"\"\"\u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52\"\"\"\n    # \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if n == 1 or n == 2:\n        return n - 1\n    # \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    res = fib(n - 1) + fib(n - 2)\n    # \u8fd4\u56de\u7ed3\u679c f(n)\n    return res\n
recursion.cpp
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nint fib(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if (n == 1 || n == 2)\n        return n - 1;\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    int res = fib(n - 1) + fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res;\n}\n
recursion.java
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nint fib(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if (n == 1 || n == 2)\n        return n - 1;\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    int res = fib(n - 1) + fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res;\n}\n
recursion.cs
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nint Fib(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if (n == 1 || n == 2)\n        return n - 1;\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    int res = Fib(n - 1) + Fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res;\n}\n
recursion.go
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nfunc fib(n int) int {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if n == 1 || n == 2 {\n        return n - 1\n    }\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    res := fib(n-1) + fib(n-2)\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res\n}\n
recursion.swift
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nfunc fib(n: Int) -> Int {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if n == 1 || n == 2 {\n        return n - 1\n    }\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    let res = fib(n: n - 1) + fib(n: n - 2)\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res\n}\n
recursion.js
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nfunction fib(n) {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if (n === 1 || n === 2) return n - 1;\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    const res = fib(n - 1) + fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res;\n}\n
recursion.ts
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nfunction fib(n: number): number {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if (n === 1 || n === 2) return n - 1;\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    const res = fib(n - 1) + fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res;\n}\n
recursion.dart
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nint fib(int n) {\n  // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n  if (n == 1 || n == 2) return n - 1;\n  // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n  int res = fib(n - 1) + fib(n - 2);\n  // \u8fd4\u56de\u7ed3\u679c f(n)\n  return res;\n}\n
recursion.rs
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nfn fib(n: i32) -> i32 {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if n == 1 || n == 2 {\n        return n - 1;\n    }\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    let res = fib(n - 1) + fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c\n    res\n}\n
recursion.c
/* \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a\u9012\u5f52 */\nint fib(int n) {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if (n == 1 || n == 2)\n        return n - 1;\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    int res = fib(n - 1) + fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res;\n}\n
recursion.zig
// \u6590\u6ce2\u90a3\u5951\u6570\u5217\nfn fib(n: i32) i32 {\n    // \u7ec8\u6b62\u6761\u4ef6 f(1) = 0, f(2) = 1\n    if (n == 1 or n == 2) {\n        return n - 1;\n    }\n    // \u9012\u5f52\u8c03\u7528 f(n) = f(n-1) + f(n-2)\n    var res: i32 = fib(n - 1) + fib(n - 2);\n    // \u8fd4\u56de\u7ed3\u679c f(n)\n    return res;\n}\n

Observing the above code, we see that it recursively calls two functions within itself, meaning that one call generates two branching calls. As illustrated below, this continuous recursive calling eventually creates a \"recursion tree\" with a depth of \\(n\\).

Figure 2-6 \u00a0 Fibonacci Sequence Recursion Tree

Fundamentally, recursion embodies the paradigm of \"breaking down a problem into smaller sub-problems.\" This divide-and-conquer strategy is crucial.

"},{"location":"chapter_computational_complexity/iteration_and_recursion/#223-comparison","title":"2.2.3 \u00a0 Comparison","text":"

Summarizing the above content, the following table shows the differences between iteration and recursion in terms of implementation, performance, and applicability.

Table: Comparison of Iteration and Recursion Characteristics

Iteration Recursion Approach Loop structure Function calls itself Time Efficiency Generally higher efficiency, no function call overhead Each function call generates overhead Memory Usage Typically uses a fixed size of memory space Accumulative function calls can use a substantial amount of stack frame space Suitable Problems Suitable for simple loop tasks, intuitive and readable code Suitable for problem decomposition, like trees, graphs, divide-and-conquer, backtracking, etc., concise and clear code structure

Tip

If you find the following content difficult to understand, consider revisiting it after reading the \"Stack\" chapter.

So, what is the intrinsic connection between iteration and recursion? Taking the above recursive function as an example, the summation operation occurs during the recursion's \"return\" phase. This means that the initially called function is actually the last to complete its summation operation, mirroring the \"last in, first out\" principle of a stack.

In fact, recursive terms like \"call stack\" and \"stack frame space\" hint at the close relationship between recursion and stacks.

  1. Recursion: When a function is called, the system allocates a new stack frame on the \"call stack\" for that function, storing local variables, parameters, return addresses, and other data.
  2. Return: When a function completes execution and returns, the corresponding stack frame is removed from the \"call stack,\" restoring the execution environment of the previous function.

Therefore, we can use an explicit stack to simulate the behavior of the call stack, thus transforming recursion into an iterative form:

PythonC++JavaC#GoSwiftJSTSDartRustCZig recursion.py
def for_loop_recur(n: int) -> int:\n    \"\"\"\u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52\"\"\"\n    # \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    stack = []\n    res = 0\n    # \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for i in range(n, 0, -1):\n        # \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.append(i)\n    # \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while stack:\n        # \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.pop()\n    # res = 1+2+3+...+n\n    return res\n
recursion.cpp
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nint forLoopRecur(int n) {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    stack<int> stack;\n    int res = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for (int i = n; i > 0; i--) {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.push(i);\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while (!stack.empty()) {\n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.top();\n        stack.pop();\n    }\n    // res = 1+2+3+...+n\n    return res;\n}\n
recursion.java
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nint forLoopRecur(int n) {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    Stack<Integer> stack = new Stack<>();\n    int res = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for (int i = n; i > 0; i--) {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.push(i);\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while (!stack.isEmpty()) {\n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.pop();\n    }\n    // res = 1+2+3+...+n\n    return res;\n}\n
recursion.cs
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nint ForLoopRecur(int n) {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    Stack<int> stack = new();\n    int res = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for (int i = n; i > 0; i--) {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.Push(i);\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while (stack.Count > 0) {\n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.Pop();\n    }\n    // res = 1+2+3+...+n\n    return res;\n}\n
recursion.go
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nfunc forLoopRecur(n int) int {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    stack := list.New()\n    res := 0\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for i := n; i > 0; i-- {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.PushBack(i)\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    for stack.Len() != 0 {\n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.Back().Value.(int)\n        stack.Remove(stack.Back())\n    }\n    // res = 1+2+3+...+n\n    return res\n}\n
recursion.swift
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nfunc forLoopRecur(n: Int) -> Int {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    var stack: [Int] = []\n    var res = 0\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for i in stride(from: n, to: 0, by: -1) {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.append(i)\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while !stack.isEmpty {\n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.removeLast()\n    }\n    // res = 1+2+3+...+n\n    return res\n}\n
recursion.js
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nfunction forLoopRecur(n) {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    const stack = [];\n    let res = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for (let i = 1; i <= n; i++) {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.push(i);\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while (stack.length) { \n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.pop();\n    }\n    // res = 1+2+3+...+n\n    return res;\n}\n
recursion.ts
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nfunction forLoopRecur(n: number): number {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808 \n    const stack: number[] = [];\n    let res: number = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for (let i = 1; i <= n; i++) {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.push(i);\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while (stack.length) { \n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.pop();\n    }\n    // res = 1+2+3+...+n\n    return res;\n}\n
recursion.dart
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nint forLoopRecur(int n) {\n  // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n  List<int> stack = [];\n  int res = 0;\n  // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n  for (int i = n; i > 0; i--) {\n    // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n    stack.add(i);\n  }\n  // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n  while (!stack.isEmpty) {\n    // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n    res += stack.removeLast();\n  }\n  // res = 1+2+3+...+n\n  return res;\n}\n
recursion.rs
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nfn for_loop_recur(n: i32) -> i32 {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    let mut stack = Vec::new();\n    let mut res = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for i in (1..=n).rev() {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack.push(i);\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while !stack.is_empty() {\n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack.pop().unwrap();\n    }\n    // res = 1+2+3+...+n\n    res\n}\n
recursion.c
/* \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52 */\nint forLoopRecur(int n) {\n    int stack[1000]; // \u501f\u52a9\u4e00\u4e2a\u5927\u6570\u7ec4\u6765\u6a21\u62df\u6808\n    int top = -1;    // \u6808\u9876\u7d22\u5f15\n    int res = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    for (int i = n; i > 0; i--) {\n        // \u901a\u8fc7\u201c\u5165\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u9012\u201d\n        stack[1 + top++] = i;\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    while (top >= 0) {\n        // \u901a\u8fc7\u201c\u51fa\u6808\u64cd\u4f5c\u201d\u6a21\u62df\u201c\u5f52\u201d\n        res += stack[top--];\n    }\n    // res = 1+2+3+...+n\n    return res;\n}\n
recursion.zig
// \u4f7f\u7528\u8fed\u4ee3\u6a21\u62df\u9012\u5f52\nfn forLoopRecur(comptime n: i32) i32 {\n    // \u4f7f\u7528\u4e00\u4e2a\u663e\u5f0f\u7684\u6808\u6765\u6a21\u62df\u7cfb\u7edf\u8c03\u7528\u6808\n    var stack: [n]i32 = undefined;\n    var res: i32 = 0;\n    // \u9012\uff1a\u9012\u5f52\u8c03\u7528\n    var i: usize = n;\n    while (i > 0) {\n        stack[i - 1] = @intCast(i);\n        i -= 1;\n    }\n    // \u5f52\uff1a\u8fd4\u56de\u7ed3\u679c\n    var index: usize = n;\n    while (index > 0) {\n        index -= 1;\n        res += stack[index];\n    }\n    // res = 1+2+3+...+n\n    return res;\n}\n

Observing the above code, when recursion is transformed into iteration, the code becomes more complex. Although iteration and recursion can often be transformed into each other, it's not always advisable to do so for two reasons:

In summary, choosing between iteration and recursion depends on the nature of the specific problem. In programming practice, weighing the pros and cons of each and choosing the appropriate method for the situation is essential.

"},{"location":"chapter_computational_complexity/performance_evaluation/","title":"2.1 \u00a0 Algorithm Efficiency Assessment","text":"

In algorithm design, we pursue the following two objectives in sequence.

  1. Finding a Solution to the Problem: The algorithm should reliably find the correct solution within the stipulated range of inputs.
  2. Seeking the Optimal Solution: For the same problem, multiple solutions might exist, and we aim to find the most efficient algorithm possible.

In other words, under the premise of being able to solve the problem, algorithm efficiency has become the main criterion for evaluating the merits of an algorithm, which includes the following two dimensions.

In short, our goal is to design data structures and algorithms that are both fast and memory-efficient. Effectively assessing algorithm efficiency is crucial because only then can we compare various algorithms and guide the process of algorithm design and optimization.

There are mainly two methods of efficiency assessment: actual testing and theoretical estimation.

"},{"location":"chapter_computational_complexity/performance_evaluation/#211-actual-testing","title":"2.1.1 \u00a0 Actual Testing","text":"

Suppose we have algorithms A and B, both capable of solving the same problem, and we need to compare their efficiencies. The most direct method is to use a computer to run these two algorithms and monitor and record their runtime and memory usage. This assessment method reflects the actual situation but has significant limitations.

On one hand, it's difficult to eliminate interference from the testing environment. Hardware configurations can affect algorithm performance. For example, algorithm A might run faster than B on one computer, but the opposite result may occur on another computer with different configurations. This means we would need to test on a variety of machines to calculate average efficiency, which is impractical.

On the other hand, conducting a full test is very resource-intensive. As the volume of input data changes, the efficiency of the algorithms may vary. For example, with smaller data volumes, algorithm A might run faster than B, but the opposite might be true with larger data volumes. Therefore, to draw convincing conclusions, we need to test a wide range of input data sizes, which requires significant computational resources.

"},{"location":"chapter_computational_complexity/performance_evaluation/#212-theoretical-estimation","title":"2.1.2 \u00a0 Theoretical Estimation","text":"

Due to the significant limitations of actual testing, we can consider evaluating algorithm efficiency solely through calculations. This estimation method is known as \"asymptotic complexity analysis,\" or simply \"complexity analysis.\"

Complexity analysis reflects the relationship between the time and space resources required for algorithm execution and the size of the input data. It describes the trend of growth in the time and space required by the algorithm as the size of the input data increases. This definition might sound complex, but we can break it down into three key points to understand it better.

Complexity analysis overcomes the disadvantages of actual testing methods, reflected in the following aspects:

Tip

If you're still confused about the concept of complexity, don't worry. We will introduce it in detail in subsequent chapters.

Complexity analysis provides us with a \"ruler\" to measure the time and space resources needed to execute an algorithm and compare the efficiency between different algorithms.

Complexity is a mathematical concept and may be abstract and challenging for beginners. From this perspective, complexity analysis might not be the best content to introduce first. However, when discussing the characteristics of a particular data structure or algorithm, it's hard to avoid analyzing its speed and space usage.

In summary, it's recommended that you establish a preliminary understanding of complexity analysis before diving deep into data structures and algorithms, so that you can carry out simple complexity analyses of algorithms.

"},{"location":"chapter_computational_complexity/space_complexity/","title":"2.4 \u00a0 Space Complexity","text":"

\"Space complexity\" is used to measure the growth trend of the memory space occupied by an algorithm as the amount of data increases. This concept is very similar to time complexity, except that \"running time\" is replaced with \"occupied memory space\".

"},{"location":"chapter_computational_complexity/space_complexity/#241-space-related-to-algorithms","title":"2.4.1 \u00a0 Space Related to Algorithms","text":"

The memory space used by an algorithm during its execution mainly includes the following types.

Generally, the scope of space complexity statistics includes both \"Temporary Space\" and \"Output Space\".

Temporary space can be further divided into three parts.

When analyzing the space complexity of a program, we typically count the Temporary Data, Stack Frame Space, and Output Data, as shown in the Figure 2-15 .

Figure 2-15 \u00a0 Space Types Used in Algorithms

The relevant code is as follows:

PythonC++JavaC#GoSwiftJSTSDartRustCZig 
class Node:\n    \"\"\"Classes\"\"\"\"\n    def __init__(self, x: int):\n        self.val: int = x               # node value\n        self.next: Node | None = None   # reference to the next node\n\ndef function() -> int:\n    \"\"\"\"Functions\"\"\"\"\"\n    # Perform certain operations...\n    return 0\n\ndef algorithm(n) -> int:    # input data\n    A = 0                   # temporary data (constant, usually in uppercase)\n    b = 0                   # temporary data (variable)\n    node = Node(0)          # temporary data (object)\n    c = function()          # Stack frame space (call function)\n    return A + b + c        # output data\n
/* Structures */\nstruct Node {\n    int val;\n    Node *next;\n    Node(int x) : val(x), next(nullptr) {}\n};\n\n/* Functions */\nint func() {\n    // Perform certain operations...\n    return 0;\n}\n\nint algorithm(int n) {          // input data\n    const int a = 0;            // temporary data (constant)\n    int b = 0;                  // temporary data (variable)\n    Node* node = new Node(0);   // temporary data (object)\n    int c = func();             // stack frame space (call function)\n    return a + b + c;           // output data\n}\n
/* Classes */\nclass Node {\n    int val;\n    Node next;\n    Node(int x) { val = x; }\n}\n\n/* Functions */\nint function() {\n    // Perform certain operations...\n    return 0;\n}\n\nint algorithm(int n) {          // input data\n    final int a = 0;            // temporary data (constant)\n    int b = 0;                  // temporary data (variable)\n    Node node = new Node(0);    // temporary data (object)\n    int c = function();         // stack frame space (call function)\n    return a + b + c;           // output data\n}\n
/* Classes */\nclass Node {\n    int val;\n    Node next;\n    Node(int x) { val = x; }\n}\n\n/* Functions */\nint Function() {\n    // Perform certain operations...\n    return 0;\n}\n\nint Algorithm(int n) {  // input data\n    const int a = 0;    // temporary data (constant)\n    int b = 0;          // temporary data (variable)\n    Node node = new(0); // temporary data (object)\n    int c = Function(); // stack frame space (call function)\n    return a + b + c;   // output data\n}\n
/* Structures */\ntype node struct {\n    val  int\n    next *node\n}\n\n/* Create node structure */\nfunc newNode(val int) *node {\n    return &node{val: val}\n}\n\n/* Functions */\nfunc function() int {\n    // Perform certain operations...\n    return 0\n}\n\nfunc algorithm(n int) int { // input data\n    const a = 0             // temporary data (constant)\n    b := 0                  // temporary storage of data (variable)\n    newNode(0)              // temporary data (object)\n    c := function()         // stack frame space (call function)\n    return a + b + c        // output data\n}\n
/* Classes */\nclass Node {\n    var val: Int\n    var next: Node?\n\n    init(x: Int) {\n        val = x\n    }\n}\n\n/* Functions */\nfunc function() -> Int {\n    // Perform certain operations...\n    return 0\n}\n\nfunc algorithm(n: Int) -> Int { // input data\n    let a = 0                   // temporary data (constant)\n    var b = 0                   // temporary data (variable)\n    let node = Node(x: 0)       // temporary data (object)\n    let c = function()          // stack frame space (call function)\n    return a + b + c            // output data\n}\n
/* Classes */\nclass Node {\n    val;\n    next;\n    constructor(val) {\n        this.val = val === undefined ? 0 : val; // node value\n        this.next = null;                       // reference to the next node\n    }\n}\n\n/* Functions */\nfunction constFunc() {\n    // Perform certain operations\n    return 0;\n}\n\nfunction algorithm(n) {         // input data\n    const a = 0;                // temporary data (constant)\n    let b = 0;                  // temporary data (variable)\n    const node = new Node(0);   // temporary data (object)\n    const c = constFunc();      // Stack frame space (calling function)\n    return a + b + c;           // output data\n}\n
/* Classes */\nclass Node {\n    val: number;\n    next: Node | null;\n    constructor(val?: number) {\n        this.val = val === undefined ? 0 : val; // node value\n        this.next = null;                       // reference to the next node\n    }\n}\n\n/* Functions */\nfunction constFunc(): number {\n    // Perform certain operations\n    return 0;\n}\n\nfunction algorithm(n: number): number { // input data\n    const a = 0;                        // temporary data (constant)\n    let b = 0;                          // temporary data (variable)\n    const node = new Node(0);           // temporary data (object)\n    const c = constFunc();              // Stack frame space (calling function)\n    return a + b + c;                   // output data\n}\n
/* Classes */\nclass Node {\n  int val;\n  Node next;\n  Node(this.val, [this.next]);\n}\n\n/* Functions */\nint function() {\n  // Perform certain operations...\n  return 0;\n}\n\nint algorithm(int n) {  // input data\n  const int a = 0;      // temporary data (constant)\n  int b = 0;            // temporary data (variable)\n  Node node = Node(0);  // temporary data (object)\n  int c = function();   // stack frame space (call function)\n  return a + b + c;     // output data\n}\n
use std::rc::Rc;\nuse std::cell::RefCell;\n\n/* Structures */\nstruct Node {\n    val: i32,\n    next: Option<Rc<RefCell<Node>>>,\n}\n\n/* Creating a Node structure */\nimpl Node {\n    fn new(val: i32) -> Self {\n        Self { val: val, next: None }\n    }\n}\n\n/* Functions */\nfn function() -> i32 {     \n    // Perform certain operations...\n    return 0;\n}\n\nfn algorithm(n: i32) -> i32 {   // input data\n    const a: i32 = 0;           // temporary data (constant)\n    let mut b = 0;              // temporary data (variable)\n    let node = Node::new(0);    // temporary data (object)\n    let c = function();         // stack frame space (call function)\n    return a + b + c;           // output data\n}\n
/* Functions */\nint func() {\n    // Perform certain operations...\n    return 0;\n}\n\nint algorithm(int n) {  // input data\n    const int a = 0;    // temporary data (constant)\n    int b = 0;          // temporary data (variable)\n    int c = func();     // stack frame space (call function)\n    return a + b + c;   // output data\n}\n
\n
"},{"location":"chapter_computational_complexity/space_complexity/#242-calculation-method","title":"2.4.2 \u00a0 Calculation Method","text":"

The method for calculating space complexity is roughly similar to that of time complexity, with the only change being the shift of the statistical object from \"number of operations\" to \"size of used space\".

However, unlike time complexity, we usually only focus on the worst-case space complexity. This is because memory space is a hard requirement, and we must ensure that there is enough memory space reserved under all input data.

Consider the following code, the term \"worst-case\" in worst-case space complexity has two meanings.

  1. Based on the worst input data: When \\(n < 10\\), the space complexity is \\(O(1)\\); but when \\(n > 10\\), the initialized array nums occupies \\(O(n)\\) space, thus the worst-case space complexity is \\(O(n)\\).
  2. Based on the peak memory used during the algorithm's execution: For example, before executing the last line, the program occupies \\(O(1)\\) space; when initializing the array nums, the program occupies \\(O(n)\\) space, hence the worst-case space complexity is \\(O(n)\\).
PythonC++JavaC#GoSwiftJSTSDartRustCZig 
def algorithm(n: int):\n    a = 0               # O(1)\n    b = [0] * 10000     # O(1)\n    if n > 10:\n        nums = [0] * n  # O(n)\n
void algorithm(int n) {\n    int a = 0;               // O(1)\n    vector<int> b(10000);    // O(1)\n    if (n > 10)\n        vector<int> nums(n); // O(n)\n}\n
void algorithm(int n) {\n    int a = 0;                   // O(1)\n    int[] b = new int[10000];    // O(1)\n    if (n > 10)\n        int[] nums = new int[n]; // O(n)\n}\n
void Algorithm(int n) {\n    int a = 0;                   // O(1)\n    int[] b = new int[10000];    // O(1)\n    if (n > 10) {\n        int[] nums = new int[n]; // O(n)\n    }\n}\n
func algorithm(n int) {\n    a := 0                      // O(1)\n    b := make([]int, 10000)     // O(1)\n    var nums []int\n    if n > 10 {\n        nums := make([]int, n)  // O(n)\n    }\n    fmt.Println(a, b, nums)\n}\n
func algorithm(n: Int) {\n    let a = 0 // O(1)\n    let b = Array(repeating: 0, count: 10000) // O(1)\n    if n > 10 {\n        let nums = Array(repeating: 0, count: n) // O(n)\n    }\n}\n
function algorithm(n) {\n    const a = 0;                   // O(1)\n    const b = new Array(10000);    // O(1)\n    if (n > 10) {\n        const nums = new Array(n); // O(n)\n    }\n}\n
function algorithm(n: number): void {\n    const a = 0;                   // O(1)\n    const b = new Array(10000);    // O(1)\n    if (n > 10) {\n        const nums = new Array(n); // O(n)\n    }\n}\n
void algorithm(int n) {\n  int a = 0;                            // O(1)\n  List<int> b = List.filled(10000, 0);  // O(1)\n  if (n > 10) {\n    List<int> nums = List.filled(n, 0); // O(n)\n  }\n}\n
fn algorithm(n: i32) {\n    let a = 0;                              // O(1)\n    let b = [0; 10000];                     // O(1)\n    if n > 10 {\n        let nums = vec![0; n as usize];     // O(n)\n    }\n}\n
void algorithm(int n) {\n    int a = 0;               // O(1)\n    int b[10000];            // O(1)\n    if (n > 10)\n        int nums[n] = {0};   // O(n)\n}\n
\n

In recursive functions, stack frame space must be taken into count. Consider the following code:

PythonC++JavaC#GoSwiftJSTSDartRustCZig 
def function() -> int:\n    # Perform certain operations\n    return 0\n\ndef loop(n: int):\n    \"\"\"Loop O(1)\"\"\"\"\"\n    for _ in range(n):\n        function()\n\ndef recur(n: int) -> int:\n    \"\"\"Recursion O(n)\"\"\"\"\"\n    if n == 1: return\n    return recur(n - 1)\n
int func() {\n    // Perform certain operations\n    return 0;\n}\n/* Cycle O(1) */\nvoid loop(int n) {\n    for (int i = 0; i < n; i++) {\n        func();\n    }\n}\n/* Recursion O(n) */\nvoid recur(int n) {\n    if (n == 1) return;\n    return recur(n - 1);\n}\n
int function() {\n    // Perform certain operations\n    return 0;\n}\n/* Cycle O(1) */\nvoid loop(int n) {\n    for (int i = 0; i < n; i++) {\n        function();\n    }\n}\n/* Recursion O(n) */\nvoid recur(int n) {\n    if (n == 1) return;\n    return recur(n - 1);\n}\n
int Function() {\n    // Perform certain operations\n    return 0;\n}\n/* Cycle O(1) */\nvoid Loop(int n) {\n    for (int i = 0; i < n; i++) {\n        Function();\n    }\n}\n/* Recursion O(n) */\nint Recur(int n) {\n    if (n == 1) return 1;\n    return Recur(n - 1);\n}\n
func function() int {\n    // Perform certain operations\n    return 0\n}\n\n/* Cycle O(1) */\nfunc loop(n int) {\n    for i := 0; i < n; i++ {\n        function()\n    }\n}\n\n/* Recursion O(n) */\nfunc recur(n int) {\n    if n == 1 {\n        return\n    }\n    recur(n - 1)\n}\n
@discardableResult\nfunc function() -> Int {\n    // Perform certain operations\n    return 0\n}\n\n/* Cycle O(1) */\nfunc loop(n: Int) {\n    for _ in 0 ..< n {\n        function()\n    }\n}\n\n/* Recursion O(n) */\nfunc recur(n: Int) {\n    if n == 1 {\n        return\n    }\n    recur(n: n - 1)\n}\n
function constFunc() {\n    // Perform certain operations\n    return 0;\n}\n/* Cycle O(1) */\nfunction loop(n) {\n    for (let i = 0; i < n; i++) {\n        constFunc();\n    }\n}\n/* Recursion O(n) */\nfunction recur(n) {\n    if (n === 1) return;\n    return recur(n - 1);\n}\n
function constFunc(): number {\n    // Perform certain operations\n    return 0;\n}\n/* Cycle O(1) */\nfunction loop(n: number): void {\n    for (let i = 0; i < n; i++) {\n        constFunc();\n    }\n}\n/* Recursion O(n) */\nfunction recur(n: number): void {\n    if (n === 1) return;\n    return recur(n - 1);\n}\n
int function() {\n  // Perform certain operations\n  return 0;\n}\n/* Cycle O(1) */\nvoid loop(int n) {\n  for (int i = 0; i < n; i++) {\n    function();\n  }\n}\n/* Recursion O(n) */\nvoid recur(int n) {\n  if (n == 1) return;\n  return recur(n - 1);\n}\n
fn function() -> i32 {\n    // Perform certain operations\n    return 0;\n}\n/* Cycle O(1) */\nfn loop(n: i32) {\n    for i in 0..n {\n        function();\n    }\n}\n/* Recursion O(n) */\nvoid recur(n: i32) {\n    if n == 1 {\n        return;\n    }\n    recur(n - 1);\n}\n
int func() {\n    // Perform certain operations\n    return 0;\n}\n/* Cycle O(1) */\nvoid loop(int n) {\n    for (int i = 0; i < n; i++) {\n        func();\n    }\n}\n/* Recursion O(n) */\nvoid recur(int n) {\n    if (n == 1) return;\n    return recur(n - 1);\n}\n
\n

The time complexity of both loop() and recur() functions is \\(O(n)\\), but their space complexities differ.

"},{"location":"chapter_computational_complexity/space_complexity/#243-common-types","title":"2.4.3 \u00a0 Common Types","text":"

Let the size of the input data be \\(n\\), the following chart displays common types of space complexities (arranged from low to high).

\\[ \\begin{aligned} O(1) < O(\\log n) < O(n) < O(n^2) < O(2^n) \\newline \\text{Constant Order} < \\text{Logarithmic Order} < \\text{Linear Order} < \\text{Quadratic Order} < \\text{Exponential Order} \\end{aligned} \\]

Figure 2-16 \u00a0 Common Types of Space Complexity

"},{"location":"chapter_computational_complexity/space_complexity/#1-constant-order-o1","title":"1. \u00a0 Constant Order \\(O(1)\\)","text":"

Constant order is common in constants, variables, objects that are independent of the size of input data \\(n\\).

Note that memory occupied by initializing variables or calling functions in a loop, which is released upon entering the next cycle, does not accumulate over space, thus the space complexity remains \\(O(1)\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig space_complexity.py
def function() -> int:\n    \"\"\"\u51fd\u6570\"\"\"\n    # \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0\n\ndef constant(n: int):\n    \"\"\"\u5e38\u6570\u9636\"\"\"\n    # \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    a = 0\n    nums = [0] * 10000\n    node = ListNode(0)\n    # \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for _ in range(n):\n        c = 0\n    # \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for _ in range(n):\n        function()\n
space_complexity.cpp
/* \u51fd\u6570 */\nint func() {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n/* \u5e38\u6570\u9636 */\nvoid constant(int n) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    const int a = 0;\n    int b = 0;\n    vector<int> nums(10000);\n    ListNode node(0);\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        int c = 0;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        func();\n    }\n}\n
space_complexity.java
/* \u51fd\u6570 */\nint function() {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n/* \u5e38\u6570\u9636 */\nvoid constant(int n) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    final int a = 0;\n    int b = 0;\n    int[] nums = new int[10000];\n    ListNode node = new ListNode(0);\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        int c = 0;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        function();\n    }\n}\n
space_complexity.cs
/* \u51fd\u6570 */\nint Function() {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n/* \u5e38\u6570\u9636 */\nvoid Constant(int n) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    int a = 0;\n    int b = 0;\n    int[] nums = new int[10000];\n    ListNode node = new(0);\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        int c = 0;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        Function();\n    }\n}\n
space_complexity.go
/* \u51fd\u6570 */\nfunc function() int {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c...\n    return 0\n}\n\n/* \u5e38\u6570\u9636 */\nfunc spaceConstant(n int) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    const a = 0\n    b := 0\n    nums := make([]int, 10000)\n    ListNode := newNode(0)\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    var c int\n    for i := 0; i < n; i++ {\n        c = 0\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for i := 0; i < n; i++ {\n        function()\n    }\n    fmt.Println(a, b, nums, c, ListNode)\n}\n
space_complexity.swift
/* \u51fd\u6570 */\n@discardableResult\nfunc function() -> Int {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0\n}\n\n/* \u5e38\u6570\u9636 */\nfunc constant(n: Int) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    let a = 0\n    var b = 0\n    let nums = Array(repeating: 0, count: 10000)\n    let node = ListNode(x: 0)\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for _ in 0 ..< n {\n        let c = 0\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for _ in 0 ..< n {\n        function()\n    }\n}\n
space_complexity.js
/* \u51fd\u6570 */\nfunction constFunc() {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n/* \u5e38\u6570\u9636 */\nfunction constant(n) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    const a = 0;\n    const b = 0;\n    const nums = new Array(10000);\n    const node = new ListNode(0);\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for (let i = 0; i < n; i++) {\n        const c = 0;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for (let i = 0; i < n; i++) {\n        constFunc();\n    }\n}\n
space_complexity.ts
/* \u51fd\u6570 */\nfunction constFunc(): number {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n/* \u5e38\u6570\u9636 */\nfunction constant(n: number): void {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    const a = 0;\n    const b = 0;\n    const nums = new Array(10000);\n    const node = new ListNode(0);\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for (let i = 0; i < n; i++) {\n        const c = 0;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for (let i = 0; i < n; i++) {\n        constFunc();\n    }\n}\n
space_complexity.dart
/* \u51fd\u6570 */\nint function() {\n  // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n  return 0;\n}\n\n/* \u5e38\u6570\u9636 */\nvoid constant(int n) {\n  // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n  final int a = 0;\n  int b = 0;\n  List<int> nums = List.filled(10000, 0);\n  ListNode node = ListNode(0);\n  // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n  for (var i = 0; i < n; i++) {\n    int c = 0;\n  }\n  // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n  for (var i = 0; i < n; i++) {\n    function();\n  }\n}\n
space_complexity.rs
/* \u51fd\u6570 */\nfn function() ->i32 {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n/* \u5e38\u6570\u9636 */\n#[allow(unused)]\nfn constant(n: i32) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    const A: i32 = 0;\n    let b = 0;\n    let nums = vec![0; 10000];\n    let node = ListNode::new(0);\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for i in 0..n {\n        let c = 0;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for i in 0..n {\n        function();\n    }\n}\n
space_complexity.c
/* \u51fd\u6570 */\nint func() {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n/* \u5e38\u6570\u9636 */\nvoid constant(int n) {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    const int a = 0;\n    int b = 0;\n    int nums[1000];\n    ListNode *node = newListNode(0);\n    free(node);\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        int c = 0;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    for (int i = 0; i < n; i++) {\n        func();\n    }\n}\n
space_complexity.zig
// \u51fd\u6570\nfn function() i32 {\n    // \u6267\u884c\u67d0\u4e9b\u64cd\u4f5c\n    return 0;\n}\n\n// \u5e38\u6570\u9636\nfn constant(n: i32) void {\n    // \u5e38\u91cf\u3001\u53d8\u91cf\u3001\u5bf9\u8c61\u5360\u7528 O(1) \u7a7a\u95f4\n    const a: i32 = 0;\n    var b: i32 = 0;\n    var nums = [_]i32{0}**10000;\n    var node = inc.ListNode(i32){.val = 0};\n    var i: i32 = 0;\n    // \u5faa\u73af\u4e2d\u7684\u53d8\u91cf\u5360\u7528 O(1) \u7a7a\u95f4\n    while (i < n) : (i += 1) {\n        var c: i32 = 0;\n        _ = c;\n    }\n    // \u5faa\u73af\u4e2d\u7684\u51fd\u6570\u5360\u7528 O(1) \u7a7a\u95f4\n    i = 0;\n    while (i < n) : (i += 1) {\n        _ = function();\n    }\n    _ = a;\n    _ = b;\n    _ = nums;\n    _ = node;\n}\n
"},{"location":"chapter_computational_complexity/space_complexity/#2-linear-order-on","title":"2. \u00a0 Linear Order \\(O(n)\\)","text":"

Linear order is common in arrays, linked lists, stacks, queues, etc., where the number of elements is proportional to \\(n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig space_complexity.py
def linear(n: int):\n    \"\"\"\u7ebf\u6027\u9636\"\"\"\n    # \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    nums = [0] * n\n    # \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    hmap = dict[int, str]()\n    for i in range(n):\n        hmap[i] = str(i)\n
space_complexity.cpp
/* \u7ebf\u6027\u9636 */\nvoid linear(int n) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    vector<int> nums(n);\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    vector<ListNode> nodes;\n    for (int i = 0; i < n; i++) {\n        nodes.push_back(ListNode(i));\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    unordered_map<int, string> map;\n    for (int i = 0; i < n; i++) {\n        map[i] = to_string(i);\n    }\n}\n
space_complexity.java
/* \u7ebf\u6027\u9636 */\nvoid linear(int n) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    int[] nums = new int[n];\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    List<ListNode> nodes = new ArrayList<>();\n    for (int i = 0; i < n; i++) {\n        nodes.add(new ListNode(i));\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    Map<Integer, String> map = new HashMap<>();\n    for (int i = 0; i < n; i++) {\n        map.put(i, String.valueOf(i));\n    }\n}\n
space_complexity.cs
/* \u7ebf\u6027\u9636 */\nvoid Linear(int n) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    int[] nums = new int[n];\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    List<ListNode> nodes = [];\n    for (int i = 0; i < n; i++) {\n        nodes.Add(new ListNode(i));\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    Dictionary<int, string> map = [];\n    for (int i = 0; i < n; i++) {\n        map.Add(i, i.ToString());\n    }\n}\n
space_complexity.go
/* \u7ebf\u6027\u9636 */\nfunc spaceLinear(n int) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    _ = make([]int, n)\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    var nodes []*node\n    for i := 0; i < n; i++ {\n        nodes = append(nodes, newNode(i))\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    m := make(map[int]string, n)\n    for i := 0; i < n; i++ {\n        m[i] = strconv.Itoa(i)\n    }\n}\n
space_complexity.swift
/* \u7ebf\u6027\u9636 */\nfunc linear(n: Int) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    let nums = Array(repeating: 0, count: n)\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    let nodes = (0 ..< n).map { ListNode(x: $0) }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    let map = Dictionary(uniqueKeysWithValues: (0 ..< n).map { ($0, \"\\($0)\") })\n}\n
space_complexity.js
/* \u7ebf\u6027\u9636 */\nfunction linear(n) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    const nums = new Array(n);\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    const nodes = [];\n    for (let i = 0; i < n; i++) {\n        nodes.push(new ListNode(i));\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    const map = new Map();\n    for (let i = 0; i < n; i++) {\n        map.set(i, i.toString());\n    }\n}\n
space_complexity.ts
/* \u7ebf\u6027\u9636 */\nfunction linear(n: number): void {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    const nums = new Array(n);\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    const nodes: ListNode[] = [];\n    for (let i = 0; i < n; i++) {\n        nodes.push(new ListNode(i));\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    const map = new Map();\n    for (let i = 0; i < n; i++) {\n        map.set(i, i.toString());\n    }\n}\n
space_complexity.dart
/* \u7ebf\u6027\u9636 */\nvoid linear(int n) {\n  // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n  List<int> nums = List.filled(n, 0);\n  // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n  List<ListNode> nodes = [];\n  for (var i = 0; i < n; i++) {\n    nodes.add(ListNode(i));\n  }\n  // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n  Map<int, String> map = HashMap();\n  for (var i = 0; i < n; i++) {\n    map.putIfAbsent(i, () => i.toString());\n  }\n}\n
space_complexity.rs
/* \u7ebf\u6027\u9636 */\n#[allow(unused)]\nfn linear(n: i32) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    let mut nums = vec![0; n as usize];\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    let mut nodes = Vec::new();\n    for i in 0..n {\n        nodes.push(ListNode::new(i))\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    let mut map = HashMap::new();\n    for i in 0..n {\n        map.insert(i, i.to_string());\n    }\n}\n
space_complexity.c
/* \u54c8\u5e0c\u8868 */\ntypedef struct {\n    int key;\n    int val;\n    UT_hash_handle hh; // \u57fa\u4e8e uthash.h \u5b9e\u73b0\n} HashTable;\n\n/* \u7ebf\u6027\u9636 */\nvoid linear(int n) {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    int *nums = malloc(sizeof(int) * n);\n    free(nums);\n\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    ListNode **nodes = malloc(sizeof(ListNode *) * n);\n    for (int i = 0; i < n; i++) {\n        nodes[i] = newListNode(i);\n    }\n    // \u5185\u5b58\u91ca\u653e\n    for (int i = 0; i < n; i++) {\n        free(nodes[i]);\n    }\n    free(nodes);\n\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    HashTable *h = NULL;\n    for (int i = 0; i < n; i++) {\n        HashTable *tmp = malloc(sizeof(HashTable));\n        tmp->key = i;\n        tmp->val = i;\n        HASH_ADD_INT(h, key, tmp);\n    }\n\n    // \u5185\u5b58\u91ca\u653e\n    HashTable *curr, *tmp;\n    HASH_ITER(hh, h, curr, tmp) {\n        HASH_DEL(h, curr);\n        free(curr);\n    }\n}\n
space_complexity.zig
// \u7ebf\u6027\u9636\nfn linear(comptime n: i32) !void {\n    // \u957f\u5ea6\u4e3a n \u7684\u6570\u7ec4\u5360\u7528 O(n) \u7a7a\u95f4\n    var nums = [_]i32{0}**n;\n    // \u957f\u5ea6\u4e3a n \u7684\u5217\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    var nodes = std.ArrayList(i32).init(std.heap.page_allocator);\n    defer nodes.deinit();\n    var i: i32 = 0;\n    while (i < n) : (i += 1) {\n        try nodes.append(i);\n    }\n    // \u957f\u5ea6\u4e3a n \u7684\u54c8\u5e0c\u8868\u5360\u7528 O(n) \u7a7a\u95f4\n    var map = std.AutoArrayHashMap(i32, []const u8).init(std.heap.page_allocator);\n    defer map.deinit();\n    var j: i32 = 0;\n    while (j < n) : (j += 1) {\n        const string = try std.fmt.allocPrint(std.heap.page_allocator, \"{d}\", .{j});\n        defer std.heap.page_allocator.free(string);\n        try map.put(i, string);\n    }\n    _ = nums;\n}\n

As shown below, this function's recursive depth is \\(n\\), meaning there are \\(n\\) instances of unreturned linear_recur() function, using \\(O(n)\\) size of stack frame space:

PythonC++JavaC#GoSwiftJSTSDartRustCZig space_complexity.py
def linear_recur(n: int):\n    \"\"\"\u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\"\"\"\n    print(\"\u9012\u5f52 n =\", n)\n    if n == 1:\n        return\n    linear_recur(n - 1)\n
space_complexity.cpp
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nvoid linearRecur(int n) {\n    cout << \"\u9012\u5f52 n = \" << n << endl;\n    if (n == 1)\n        return;\n    linearRecur(n - 1);\n}\n
space_complexity.java
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nvoid linearRecur(int n) {\n    System.out.println(\"\u9012\u5f52 n = \" + n);\n    if (n == 1)\n        return;\n    linearRecur(n - 1);\n}\n
space_complexity.cs
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nvoid LinearRecur(int n) {\n    Console.WriteLine(\"\u9012\u5f52 n = \" + n);\n    if (n == 1) return;\n    LinearRecur(n - 1);\n}\n
space_complexity.go
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunc spaceLinearRecur(n int) {\n    fmt.Println(\"\u9012\u5f52 n =\", n)\n    if n == 1 {\n        return\n    }\n    spaceLinearRecur(n - 1)\n}\n
space_complexity.swift
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunc linearRecur(n: Int) {\n    print(\"\u9012\u5f52 n = \\(n)\")\n    if n == 1 {\n        return\n    }\n    linearRecur(n: n - 1)\n}\n
space_complexity.js
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction linearRecur(n) {\n    console.log(`\u9012\u5f52 n = ${n}`);\n    if (n === 1) return;\n    linearRecur(n - 1);\n}\n
space_complexity.ts
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction linearRecur(n: number): void {\n    console.log(`\u9012\u5f52 n = ${n}`);\n    if (n === 1) return;\n    linearRecur(n - 1);\n}\n
space_complexity.dart
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nvoid linearRecur(int n) {\n  print('\u9012\u5f52 n = $n');\n  if (n == 1) return;\n  linearRecur(n - 1);\n}\n
space_complexity.rs
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfn linear_recur(n: i32) {\n    println!(\"\u9012\u5f52 n = {}\", n);\n    if n == 1 {return};\n    linear_recur(n - 1);\n}\n
space_complexity.c
/* \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nvoid linearRecur(int n) {\n    printf(\"\u9012\u5f52 n = %d\\r\\n\", n);\n    if (n == 1)\n        return;\n    linearRecur(n - 1);\n}\n
space_complexity.zig
// \u7ebf\u6027\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\nfn linearRecur(comptime n: i32) void {\n    std.debug.print(\"\u9012\u5f52 n = {}\\n\", .{n});\n    if (n == 1) return;\n    linearRecur(n - 1);\n}\n

Figure 2-17 \u00a0 Recursive Function Generating Linear Order Space Complexity

"},{"location":"chapter_computational_complexity/space_complexity/#3-quadratic-order-on2","title":"3. \u00a0 Quadratic Order \\(O(n^2)\\)","text":"

Quadratic order is common in matrices and graphs, where the number of elements is quadratic to \\(n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig space_complexity.py
def quadratic(n: int):\n    \"\"\"\u5e73\u65b9\u9636\"\"\"\n    # \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    num_matrix = [[0] * n for _ in range(n)]\n
space_complexity.cpp
/* \u5e73\u65b9\u9636 */\nvoid quadratic(int n) {\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    vector<vector<int>> numMatrix;\n    for (int i = 0; i < n; i++) {\n        vector<int> tmp;\n        for (int j = 0; j < n; j++) {\n            tmp.push_back(0);\n        }\n        numMatrix.push_back(tmp);\n    }\n}\n
space_complexity.java
/* \u5e73\u65b9\u9636 */\nvoid quadratic(int n) {\n    // \u77e9\u9635\u5360\u7528 O(n^2) \u7a7a\u95f4\n    int[][] numMatrix = new int[n][n];\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    List<List<Integer>> numList = new ArrayList<>();\n    for (int i = 0; i < n; i++) {\n        List<Integer> tmp = new ArrayList<>();\n        for (int j = 0; j < n; j++) {\n            tmp.add(0);\n        }\n        numList.add(tmp);\n    }\n}\n
space_complexity.cs
/* \u5e73\u65b9\u9636 */\nvoid Quadratic(int n) {\n    // \u77e9\u9635\u5360\u7528 O(n^2) \u7a7a\u95f4\n    int[,] numMatrix = new int[n, n];\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    List<List<int>> numList = [];\n    for (int i = 0; i < n; i++) {\n        List<int> tmp = [];\n        for (int j = 0; j < n; j++) {\n            tmp.Add(0);\n        }\n        numList.Add(tmp);\n    }\n}\n
space_complexity.go
/* \u5e73\u65b9\u9636 */\nfunc spaceQuadratic(n int) {\n    // \u77e9\u9635\u5360\u7528 O(n^2) \u7a7a\u95f4\n    numMatrix := make([][]int, n)\n    for i := 0; i < n; i++ {\n        numMatrix[i] = make([]int, n)\n    }\n}\n
space_complexity.swift
/* \u5e73\u65b9\u9636 */\nfunc quadratic(n: Int) {\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    let numList = Array(repeating: Array(repeating: 0, count: n), count: n)\n}\n
space_complexity.js
/* \u5e73\u65b9\u9636 */\nfunction quadratic(n) {\n    // \u77e9\u9635\u5360\u7528 O(n^2) \u7a7a\u95f4\n    const numMatrix = Array(n)\n        .fill(null)\n        .map(() => Array(n).fill(null));\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    const numList = [];\n    for (let i = 0; i < n; i++) {\n        const tmp = [];\n        for (let j = 0; j < n; j++) {\n            tmp.push(0);\n        }\n        numList.push(tmp);\n    }\n}\n
space_complexity.ts
/* \u5e73\u65b9\u9636 */\nfunction quadratic(n: number): void {\n    // \u77e9\u9635\u5360\u7528 O(n^2) \u7a7a\u95f4\n    const numMatrix = Array(n)\n        .fill(null)\n        .map(() => Array(n).fill(null));\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    const numList = [];\n    for (let i = 0; i < n; i++) {\n        const tmp = [];\n        for (let j = 0; j < n; j++) {\n            tmp.push(0);\n        }\n        numList.push(tmp);\n    }\n}\n
space_complexity.dart
/* \u5e73\u65b9\u9636 */\nvoid quadratic(int n) {\n  // \u77e9\u9635\u5360\u7528 O(n^2) \u7a7a\u95f4\n  List<List<int>> numMatrix = List.generate(n, (_) => List.filled(n, 0));\n  // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n  List<List<int>> numList = [];\n  for (var i = 0; i < n; i++) {\n    List<int> tmp = [];\n    for (int j = 0; j < n; j++) {\n      tmp.add(0);\n    }\n    numList.add(tmp);\n  }\n}\n
space_complexity.rs
/* \u5e73\u65b9\u9636 */\n#[allow(unused)]\nfn quadratic(n: i32) {\n    // \u77e9\u9635\u5360\u7528 O(n^2) \u7a7a\u95f4\n    let num_matrix = vec![vec![0; n as usize]; n as usize];\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    let mut num_list = Vec::new();\n    for i in 0..n {\n        let mut tmp = Vec::new();\n        for j in 0..n {\n            tmp.push(0);\n        }\n        num_list.push(tmp);\n    }\n}\n
space_complexity.c
/* \u5e73\u65b9\u9636 */\nvoid quadratic(int n) {\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    int **numMatrix = malloc(sizeof(int *) * n);\n    for (int i = 0; i < n; i++) {\n        int *tmp = malloc(sizeof(int) * n);\n        for (int j = 0; j < n; j++) {\n            tmp[j] = 0;\n        }\n        numMatrix[i] = tmp;\n    }\n\n    // \u5185\u5b58\u91ca\u653e\n    for (int i = 0; i < n; i++) {\n        free(numMatrix[i]);\n    }\n    free(numMatrix);\n}\n
space_complexity.zig
// \u5e73\u65b9\u9636\nfn quadratic(n: i32) !void {\n    // \u4e8c\u7ef4\u5217\u8868\u5360\u7528 O(n^2) \u7a7a\u95f4\n    var nodes = std.ArrayList(std.ArrayList(i32)).init(std.heap.page_allocator);\n    defer nodes.deinit();\n    var i: i32 = 0;\n    while (i < n) : (i += 1) {\n        var tmp = std.ArrayList(i32).init(std.heap.page_allocator);\n        defer tmp.deinit();\n        var j: i32 = 0;\n        while (j < n) : (j += 1) {\n            try tmp.append(0);\n        }\n        try nodes.append(tmp);\n    }\n}\n

As shown below, the recursive depth of this function is \\(n\\), and in each recursive call, an array is initialized with lengths \\(n\\), \\(n-1\\), \\(\\dots\\), \\(2\\), \\(1\\), averaging \\(n/2\\), thus overall occupying \\(O(n^2)\\) space:

PythonC++JavaC#GoSwiftJSTSDartRustCZig space_complexity.py
def quadratic_recur(n: int) -> int:\n    \"\"\"\u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\"\"\"\n    if n <= 0:\n        return 0\n    # \u6570\u7ec4 nums \u957f\u5ea6\u4e3a n, n-1, ..., 2, 1\n    nums = [0] * n\n    return quadratic_recur(n - 1)\n
space_complexity.cpp
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint quadraticRecur(int n) {\n    if (n <= 0)\n        return 0;\n    vector<int> nums(n);\n    cout << \"\u9012\u5f52 n = \" << n << \" \u4e2d\u7684 nums \u957f\u5ea6 = \" << nums.size() << endl;\n    return quadraticRecur(n - 1);\n}\n
space_complexity.java
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint quadraticRecur(int n) {\n    if (n <= 0)\n        return 0;\n    // \u6570\u7ec4 nums \u957f\u5ea6\u4e3a n, n-1, ..., 2, 1\n    int[] nums = new int[n];\n    System.out.println(\"\u9012\u5f52 n = \" + n + \" \u4e2d\u7684 nums \u957f\u5ea6 = \" + nums.length);\n    return quadraticRecur(n - 1);\n}\n
space_complexity.cs
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint QuadraticRecur(int n) {\n    if (n <= 0) return 0;\n    int[] nums = new int[n];\n    Console.WriteLine(\"\u9012\u5f52 n = \" + n + \" \u4e2d\u7684 nums \u957f\u5ea6 = \" + nums.Length);\n    return QuadraticRecur(n - 1);\n}\n
space_complexity.go
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunc spaceQuadraticRecur(n int) int {\n    if n <= 0 {\n        return 0\n    }\n    nums := make([]int, n)\n    fmt.Printf(\"\u9012\u5f52 n = %d \u4e2d\u7684 nums \u957f\u5ea6 = %d \\n\", n, len(nums))\n    return spaceQuadraticRecur(n - 1)\n}\n
space_complexity.swift
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\n@discardableResult\nfunc quadraticRecur(n: Int) -> Int {\n    if n <= 0 {\n        return 0\n    }\n    // \u6570\u7ec4 nums \u957f\u5ea6\u4e3a n, n-1, ..., 2, 1\n    let nums = Array(repeating: 0, count: n)\n    print(\"\u9012\u5f52 n = \\(n) \u4e2d\u7684 nums \u957f\u5ea6 = \\(nums.count)\")\n    return quadraticRecur(n: n - 1)\n}\n
space_complexity.js
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction quadraticRecur(n) {\n    if (n <= 0) return 0;\n    const nums = new Array(n);\n    console.log(`\u9012\u5f52 n = ${n} \u4e2d\u7684 nums \u957f\u5ea6 = ${nums.length}`);\n    return quadraticRecur(n - 1);\n}\n
space_complexity.ts
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction quadraticRecur(n: number): number {\n    if (n <= 0) return 0;\n    const nums = new Array(n);\n    console.log(`\u9012\u5f52 n = ${n} \u4e2d\u7684 nums \u957f\u5ea6 = ${nums.length}`);\n    return quadraticRecur(n - 1);\n}\n
space_complexity.dart
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint quadraticRecur(int n) {\n  if (n <= 0) return 0;\n  List<int> nums = List.filled(n, 0);\n  print('\u9012\u5f52 n = $n \u4e2d\u7684 nums \u957f\u5ea6 = ${nums.length}');\n  return quadraticRecur(n - 1);\n}\n
space_complexity.rs
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfn quadratic_recur(n: i32) -> i32 {\n    if n <= 0 {return 0};\n    // \u6570\u7ec4 nums \u957f\u5ea6\u4e3a n, n-1, ..., 2, 1\n    let nums = vec![0; n as usize];\n    println!(\"\u9012\u5f52 n = {} \u4e2d\u7684 nums \u957f\u5ea6 = {}\", n, nums.len());\n    return quadratic_recur(n - 1);\n}\n
space_complexity.c
/* \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint quadraticRecur(int n) {\n    if (n <= 0)\n        return 0;\n    int *nums = malloc(sizeof(int) * n);\n    printf(\"\u9012\u5f52 n = %d \u4e2d\u7684 nums \u957f\u5ea6 = %d\\r\\n\", n, n);\n    int res = quadraticRecur(n - 1);\n    free(nums);\n    return res;\n}\n
space_complexity.zig
// \u5e73\u65b9\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\nfn quadraticRecur(comptime n: i32) i32 {\n    if (n <= 0) return 0;\n    var nums = [_]i32{0}**n;\n    std.debug.print(\"\u9012\u5f52 n = {} \u4e2d\u7684 nums \u957f\u5ea6 = {}\\n\", .{n, nums.len});\n    return quadraticRecur(n - 1);\n}\n

Figure 2-18 \u00a0 Recursive Function Generating Quadratic Order Space Complexity

"},{"location":"chapter_computational_complexity/space_complexity/#4-exponential-order-o2n","title":"4. \u00a0 Exponential Order \\(O(2^n)\\)","text":"

Exponential order is common in binary trees. Observe the below image, a \"full binary tree\" with \\(n\\) levels has \\(2^n - 1\\) nodes, occupying \\(O(2^n)\\) space:

PythonC++JavaC#GoSwiftJSTSDartRustCZig space_complexity.py
def build_tree(n: int) -> TreeNode | None:\n    \"\"\"\u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09\"\"\"\n    if n == 0:\n        return None\n    root = TreeNode(0)\n    root.left = build_tree(n - 1)\n    root.right = build_tree(n - 1)\n    return root\n
space_complexity.cpp
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nTreeNode *buildTree(int n) {\n    if (n == 0)\n        return nullptr;\n    TreeNode *root = new TreeNode(0);\n    root->left = buildTree(n - 1);\n    root->right = buildTree(n - 1);\n    return root;\n}\n
space_complexity.java
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nTreeNode buildTree(int n) {\n    if (n == 0)\n        return null;\n    TreeNode root = new TreeNode(0);\n    root.left = buildTree(n - 1);\n    root.right = buildTree(n - 1);\n    return root;\n}\n
space_complexity.cs
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nTreeNode? BuildTree(int n) {\n    if (n == 0) return null;\n    TreeNode root = new(0) {\n        left = BuildTree(n - 1),\n        right = BuildTree(n - 1)\n    };\n    return root;\n}\n
space_complexity.go
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nfunc buildTree(n int) *treeNode {\n    if n == 0 {\n        return nil\n    }\n    root := newTreeNode(0)\n    root.left = buildTree(n - 1)\n    root.right = buildTree(n - 1)\n    return root\n}\n
space_complexity.swift
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nfunc buildTree(n: Int) -> TreeNode? {\n    if n == 0 {\n        return nil\n    }\n    let root = TreeNode(x: 0)\n    root.left = buildTree(n: n - 1)\n    root.right = buildTree(n: n - 1)\n    return root\n}\n
space_complexity.js
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nfunction buildTree(n) {\n    if (n === 0) return null;\n    const root = new TreeNode(0);\n    root.left = buildTree(n - 1);\n    root.right = buildTree(n - 1);\n    return root;\n}\n
space_complexity.ts
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nfunction buildTree(n: number): TreeNode | null {\n    if (n === 0) return null;\n    const root = new TreeNode(0);\n    root.left = buildTree(n - 1);\n    root.right = buildTree(n - 1);\n    return root;\n}\n
space_complexity.dart
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nTreeNode? buildTree(int n) {\n  if (n == 0) return null;\n  TreeNode root = TreeNode(0);\n  root.left = buildTree(n - 1);\n  root.right = buildTree(n - 1);\n  return root;\n}\n
space_complexity.rs
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nfn build_tree(n: i32) -> Option<Rc<RefCell<TreeNode>>> {\n    if n == 0 {return None};\n    let root = TreeNode::new(0);\n    root.borrow_mut().left = build_tree(n - 1);\n    root.borrow_mut().right = build_tree(n - 1);\n    return Some(root);\n}\n
space_complexity.c
/* \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09 */\nTreeNode *buildTree(int n) {\n    if (n == 0)\n        return NULL;\n    TreeNode *root = newTreeNode(0);\n    root->left = buildTree(n - 1);\n    root->right = buildTree(n - 1);\n    return root;\n}\n
space_complexity.zig
// \u6307\u6570\u9636\uff08\u5efa\u7acb\u6ee1\u4e8c\u53c9\u6811\uff09\nfn buildTree(mem_allocator: std.mem.Allocator, n: i32) !?*inc.TreeNode(i32) {\n    if (n == 0) return null;\n    const root = try mem_allocator.create(inc.TreeNode(i32));\n    root.init(0);\n    root.left = try buildTree(mem_allocator, n - 1);\n    root.right = try buildTree(mem_allocator, n - 1);\n    return root;\n}\n

Figure 2-19 \u00a0 Full Binary Tree Generating Exponential Order Space Complexity

"},{"location":"chapter_computational_complexity/space_complexity/#5-logarithmic-order-olog-n","title":"5. \u00a0 Logarithmic Order \\(O(\\log n)\\)","text":"

Logarithmic order is common in divide-and-conquer algorithms. For example, in merge sort, an array of length \\(n\\) is recursively divided in half each round, forming a recursion tree of height \\(\\log n\\), using \\(O(\\log n)\\) stack frame space.

Another example is converting a number to a string. Given a positive integer \\(n\\), its number of digits is \\(\\log_{10} n + 1\\), corresponding to the length of the string, thus the space complexity is \\(O(\\log_{10} n + 1) = O(\\log n)\\).

"},{"location":"chapter_computational_complexity/space_complexity/#244-balancing-time-and-space","title":"2.4.4 \u00a0 Balancing Time and Space","text":"

Ideally, we aim for both time complexity and space complexity to be optimal. However, in practice, optimizing both simultaneously is often difficult.

Lowering time complexity usually comes at the cost of increased space complexity, and vice versa. The approach of sacrificing memory space to improve algorithm speed is known as \"space-time tradeoff\"; the reverse is known as \"time-space tradeoff\".

The choice depends on which aspect we value more. In most cases, time is more precious than space, so \"space-time tradeoff\" is often the more common strategy. Of course, controlling space complexity is also very important when dealing with large volumes of data.

"},{"location":"chapter_computational_complexity/summary/","title":"2.5 \u00a0 Summary","text":""},{"location":"chapter_computational_complexity/summary/#1-key-review","title":"1. \u00a0 Key Review","text":"

Algorithm Efficiency Assessment

Time Complexity

Space Complexity

"},{"location":"chapter_computational_complexity/summary/#2-q-a","title":"2. \u00a0 Q & A","text":"

Is the space complexity of tail recursion \\(O(1)\\)?

Theoretically, the space complexity of a tail-recursive function can be optimized to \\(O(1)\\). However, most programming languages (such as Java, Python, C++, Go, C#) do not support automatic optimization of tail recursion, so it's generally considered to have a space complexity of \\(O(n)\\).

What is the difference between the terms 'function' and 'method'?

A \"function\" can be executed independently, with all parameters passed explicitly. A \"method\" is associated with an object and is implicitly passed to the object calling it, able to operate on the data contained within an instance of a class.

Here are some examples from common programming languages:

Does the 'Common Types of Space Complexity' figure reflect the absolute size of occupied space?

No, the figure shows space complexities, which reflect growth trends, not the absolute size of the occupied space.

If you take \\(n = 8\\), you might find that the values of each curve don't correspond to their functions. This is because each curve includes a constant term, intended to compress the value range into a visually comfortable range.

In practice, since we usually don't know the \"constant term\" complexity of each method, it's generally not possible to choose the best solution for \\(n = 8\\) based solely on complexity. However, for \\(n = 8^5\\), it's much easier to choose, as the growth trend becomes dominant.

"},{"location":"chapter_computational_complexity/time_complexity/","title":"2.3 \u00a0 Time Complexity","text":"

Time complexity is a concept used to measure how the run time of an algorithm increases with the size of the input data. Understanding time complexity is crucial for accurately assessing the efficiency of an algorithm.

  1. Determining the Running Platform: This includes hardware configuration, programming language, system environment, etc., all of which can affect the efficiency of code execution.
  2. Evaluating the Run Time for Various Computational Operations: For instance, an addition operation + might take 1 ns, a multiplication operation * might take 10 ns, a print operation print() might take 5 ns, etc.
  3. Counting All the Computational Operations in the Code: Summing the execution times of all these operations gives the total run time.

For example, consider the following code with an input size of \\(n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig 
# Under an operating platform\ndef algorithm(n: int):\n    a = 2      # 1 ns\n    a = a + 1  # 1 ns\n    a = a * 2  # 10 ns\n    # Cycle n times\n    for _ in range(n):  # 1 ns\n        print(0)        # 5 ns\n
// Under a particular operating platform\nvoid algorithm(int n) {\n    int a = 2;  // 1 ns\n    a = a + 1;  // 1 ns\n    a = a * 2;  // 10 ns\n    // Loop n times\n    for (int i = 0; i < n; i++) {  // 1 ns , every round i++ is executed\n        cout << 0 << endl;         // 5 ns\n    }\n}\n
// Under a particular operating platform\nvoid algorithm(int n) {\n    int a = 2;  // 1 ns\n    a = a + 1;  // 1 ns\n    a = a * 2;  // 10 ns\n    // Loop n times\n    for (int i = 0; i < n; i++) {  // 1 ns , every round i++ is executed\n        System.out.println(0);     // 5 ns\n    }\n}\n
// Under a particular operating platform\nvoid Algorithm(int n) {\n    int a = 2;  // 1 ns\n    a = a + 1;  // 1 ns\n    a = a * 2;  // 10 ns\n    // Loop n times\n    for (int i = 0; i < n; i++) {  // 1 ns , every round i++ is executed\n        Console.WriteLine(0);      // 5 ns\n    }\n}\n
// Under a particular operating platform\nfunc algorithm(n int) {\n    a := 2     // 1 ns\n    a = a + 1  // 1 ns\n    a = a * 2  // 10 ns\n    // Loop n times\n    for i := 0; i < n; i++ {  // 1 ns\n        fmt.Println(a)        // 5 ns\n    }\n}\n
// Under a particular operating platform\nfunc algorithm(n: Int) {\n    var a = 2 // 1 ns\n    a = a + 1 // 1 ns\n    a = a * 2 // 10 ns\n    // Loop n times\n    for _ in 0 ..< n { // 1 ns\n        print(0) // 5 ns\n    }\n}\n
// Under a particular operating platform\nfunction algorithm(n) {\n    var a = 2; // 1 ns\n    a = a + 1; // 1 ns\n    a = a * 2; // 10 ns\n    // Loop n times\n    for(let i = 0; i < n; i++) { // 1 ns , every round i++ is executed\n        console.log(0); // 5 ns\n    }\n}\n
// Under a particular operating platform\nfunction algorithm(n: number): void {\n    var a: number = 2; // 1 ns\n    a = a + 1; // 1 ns\n    a = a * 2; // 10 ns\n    // Loop n times\n    for(let i = 0; i < n; i++) { // 1 ns , every round i++ is executed\n        console.log(0); // 5 ns\n    }\n}\n
// Under a particular operating platform\nvoid algorithm(int n) {\n  int a = 2; // 1 ns\n  a = a + 1; // 1 ns\n  a = a * 2; // 10 ns\n  // Loop n times\n  for (int i = 0; i < n; i++) { // 1 ns , every round i++ is executed\n    print(0); // 5 ns\n  }\n}\n
// Under a particular operating platform\nfn algorithm(n: i32) {\n    let mut a = 2;      // 1 ns\n    a = a + 1;          // 1 ns\n    a = a * 2;          // 10 ns\n    // Loop n times\n    for _ in 0..n {     // 1 ns for each round i++\n        println!(\"{}\", 0);  // 5 ns\n    }\n}\n
// Under a particular operating platform\nvoid algorithm(int n) {\n    int a = 2;  // 1 ns\n    a = a + 1;  // 1 ns\n    a = a * 2;  // 10 ns\n    // Loop n times\n    for (int i = 0; i < n; i++) {   // 1 ns , every round i++ is executed\n        printf(\"%d\", 0);            // 5 ns\n    }\n}\n
// Under a particular operating platform\nfn algorithm(n: usize) void {\n    var a: i32 = 2; // 1 ns\n    a += 1; // 1 ns\n    a *= 2; // 10 ns\n    // Loop n times\n    for (0..n) |_| { // 1 ns\n        std.debug.print(\"{}\\n\", .{0}); // 5 ns\n    }\n}\n

Using the above method, the run time of the algorithm can be calculated as \\((6n + 12)\\) ns:

\\[ 1 + 1 + 10 + (1 + 5) \\times n = 6n + 12 \\]

However, in practice, counting the run time of an algorithm is neither practical nor reasonable. First, we don't want to tie the estimated time to the running platform, as algorithms need to run on various platforms. Second, it's challenging to know the run time for each type of operation, making the estimation process difficult.

"},{"location":"chapter_computational_complexity/time_complexity/#231-assessing-time-growth-trend","title":"2.3.1 \u00a0 Assessing Time Growth Trend","text":"

Time complexity analysis does not count the algorithm's run time, but rather the growth trend of the run time as the data volume increases.

Let's understand this concept of \"time growth trend\" with an example. Assume the input data size is \\(n\\), and consider three algorithms A, B, and C:

PythonC++JavaC#GoSwiftJSTSDartRustCZig 
# Time complexity of algorithm A: constant order\ndef algorithm_A(n: int):\n    print(0)\n# Time complexity of algorithm B: linear order\ndef algorithm_B(n: int):\n    for _ in range(n):\n        print(0)\n# Time complexity of algorithm C: constant order\ndef algorithm_C(n: int):\n    for _ in range(1000000):\n        print(0)\n
// Time complexity of algorithm A: constant order\nvoid algorithm_A(int n) {\n    cout << 0 << endl;\n}\n// Time complexity of algorithm B: linear order\nvoid algorithm_B(int n) {\n    for (int i = 0; i < n; i++) {\n        cout << 0 << endl;\n    }\n}\n// Time complexity of algorithm C: constant order\nvoid algorithm_C(int n) {\n    for (int i = 0; i < 1000000; i++) {\n        cout << 0 << endl;\n    }\n}\n
// Time complexity of algorithm A: constant order\nvoid algorithm_A(int n) {\n    System.out.println(0);\n}\n// Time complexity of algorithm B: linear order\nvoid algorithm_B(int n) {\n    for (int i = 0; i < n; i++) {\n        System.out.println(0);\n    }\n}\n// Time complexity of algorithm C: constant order\nvoid algorithm_C(int n) {\n    for (int i = 0; i < 1000000; i++) {\n        System.out.println(0);\n    }\n}\n
// Time complexity of algorithm A: constant order\nvoid AlgorithmA(int n) {\n    Console.WriteLine(0);\n}\n// Time complexity of algorithm B: linear order\nvoid AlgorithmB(int n) {\n    for (int i = 0; i < n; i++) {\n        Console.WriteLine(0);\n    }\n}\n// Time complexity of algorithm C: constant order\nvoid AlgorithmC(int n) {\n    for (int i = 0; i < 1000000; i++) {\n        Console.WriteLine(0);\n    }\n}\n
// Time complexity of algorithm A: constant order\nfunc algorithm_A(n int) {\n    fmt.Println(0)\n}\n// Time complexity of algorithm B: linear order\nfunc algorithm_B(n int) {\n    for i := 0; i < n; i++ {\n        fmt.Println(0)\n    }\n}\n// Time complexity of algorithm C: constant order\nfunc algorithm_C(n int) {\n    for i := 0; i < 1000000; i++ {\n        fmt.Println(0)\n    }\n}\n
// Time complexity of algorithm A: constant order\nfunc algorithmA(n: Int) {\n    print(0)\n}\n\n// Time complexity of algorithm B: linear order\nfunc algorithmB(n: Int) {\n    for _ in 0 ..< n {\n        print(0)\n    }\n}\n\n// Time complexity of algorithm C: constant order\nfunc algorithmC(n: Int) {\n    for _ in 0 ..< 1000000 {\n        print(0)\n    }\n}\n
// Time complexity of algorithm A: constant order\nfunction algorithm_A(n) {\n    console.log(0);\n}\n// Time complexity of algorithm B: linear order\nfunction algorithm_B(n) {\n    for (let i = 0; i < n; i++) {\n        console.log(0);\n    }\n}\n// Time complexity of algorithm C: constant order\nfunction algorithm_C(n) {\n    for (let i = 0; i < 1000000; i++) {\n        console.log(0);\n    }\n}\n
// Time complexity of algorithm A: constant order\nfunction algorithm_A(n: number): void {\n    console.log(0);\n}\n// Time complexity of algorithm B: linear order\nfunction algorithm_B(n: number): void {\n    for (let i = 0; i < n; i++) {\n        console.log(0);\n    }\n}\n// Time complexity of algorithm C: constant order\nfunction algorithm_C(n: number): void {\n    for (let i = 0; i < 1000000; i++) {\n        console.log(0);\n    }\n}\n
// Time complexity of algorithm A: constant order\nvoid algorithmA(int n) {\n  print(0);\n}\n// Time complexity of algorithm B: linear order\nvoid algorithmB(int n) {\n  for (int i = 0; i < n; i++) {\n    print(0);\n  }\n}\n// Time complexity of algorithm C: constant order\nvoid algorithmC(int n) {\n  for (int i = 0; i < 1000000; i++) {\n    print(0);\n  }\n}\n
// Time complexity of algorithm A: constant order\nfn algorithm_A(n: i32) {\n    println!(\"{}\", 0);\n}\n// Time complexity of algorithm B: linear order\nfn algorithm_B(n: i32) {\n    for _ in 0..n {\n        println!(\"{}\", 0);\n    }\n}\n// Time complexity of algorithm C: constant order\nfn algorithm_C(n: i32) {\n    for _ in 0..1000000 {\n        println!(\"{}\", 0);\n    }\n}\n
// Time complexity of algorithm A: constant order\nvoid algorithm_A(int n) {\n    printf(\"%d\", 0);\n}\n// Time complexity of algorithm B: linear order\nvoid algorithm_B(int n) {\n    for (int i = 0; i < n; i++) {\n        printf(\"%d\", 0);\n    }\n}\n// Time complexity of algorithm C: constant order\nvoid algorithm_C(int n) {\n    for (int i = 0; i < 1000000; i++) {\n        printf(\"%d\", 0);\n    }\n}\n
// Time complexity of algorithm A: constant order\nfn algorithm_A(n: usize) void {\n    _ = n;\n    std.debug.print(\"{}\\n\", .{0});\n}\n// Time complexity of algorithm B: linear order\nfn algorithm_B(n: i32) void {\n    for (0..n) |_| {\n        std.debug.print(\"{}\\n\", .{0});\n    }\n}\n// Time complexity of algorithm C: constant order\nfn algorithm_C(n: i32) void {\n    _ = n;\n    for (0..1000000) |_| {\n        std.debug.print(\"{}\\n\", .{0});\n    }\n}\n

The following figure shows the time complexities of these three algorithms.

Figure 2-7 \u00a0 Time Growth Trend of Algorithms A, B, and C

Compared to directly counting the run time of an algorithm, what are the characteristics of time complexity analysis?

"},{"location":"chapter_computational_complexity/time_complexity/#232-asymptotic-upper-bound","title":"2.3.2 \u00a0 Asymptotic Upper Bound","text":"

Consider a function with an input size of \\(n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig 
def algorithm(n: int):\n    a = 1      # +1\n    a = a + 1  # +1\n    a = a * 2  # +1\n    # Cycle n times\n    for i in range(n):  # +1\n        print(0)        # +1\n
void algorithm(int n) {\n    int a = 1;  // +1\n    a = a + 1;  // +1\n    a = a * 2;  // +1\n    // Loop n times\n    for (int i = 0; i < n; i++) { // +1 (execute i ++ every round)\n        cout << 0 << endl;    // +1\n    }\n}\n
void algorithm(int n) {\n    int a = 1;  // +1\n    a = a + 1;  // +1\n    a = a * 2;  // +1\n    // Loop n times\n    for (int i = 0; i < n; i++) { // +1 (execute i ++ every round)\n        System.out.println(0);    // +1\n    }\n}\n
void Algorithm(int n) {\n    int a = 1;  // +1\n    a = a + 1;  // +1\n    a = a * 2;  // +1\n    // Loop n times\n    for (int i = 0; i < n; i++) {   // +1 (execute i ++ every round)\n        Console.WriteLine(0);   // +1\n    }\n}\n
func algorithm(n int) {\n    a := 1      // +1\n    a = a + 1   // +1\n    a = a * 2   // +1\n    // Loop n times\n    for i := 0; i < n; i++ {   // +1\n        fmt.Println(a)         // +1\n    }\n}\n
func algorithm(n: Int) {\n    var a = 1 // +1\n    a = a + 1 // +1\n    a = a * 2 // +1\n    // Loop n times\n    for _ in 0 ..< n { // +1\n        print(0) // +1\n    }\n}\n
function algorithm(n) {\n    var a = 1; // +1\n    a += 1; // +1\n    a *= 2; // +1\n    // Loop n times\n    for(let i = 0; i < n; i++){ // +1 (execute i ++ every round)\n        console.log(0); // +1\n    }\n}\n
function algorithm(n: number): void{\n    var a: number = 1; // +1\n    a += 1; // +1\n    a *= 2; // +1\n    // Loop n times\n    for(let i = 0; i < n; i++){ // +1 (execute i ++ every round)\n        console.log(0); // +1\n    }\n}\n
void algorithm(int n) {\n  int a = 1; // +1\n  a = a + 1; // +1\n  a = a * 2; // +1\n  // Loop n times\n  for (int i = 0; i < n; i++) { // +1 (execute i ++ every round)\n    print(0); // +1\n  }\n}\n
fn algorithm(n: i32) {\n    let mut a = 1;   // +1\n    a = a + 1;      // +1\n    a = a * 2;      // +1\n\n    // Loop n times\n    for _ in 0..n { // +1 (execute i ++ every round)\n        println!(\"{}\", 0); // +1\n    }\n}\n
void algorithm(int n) {\n    int a = 1;  // +1\n    a = a + 1;  // +1\n    a = a * 2;  // +1\n    // Loop n times\n    for (int i = 0; i < n; i++) {   // +1 (execute i ++ every round)\n        printf(\"%d\", 0);            // +1\n    }\n} \n
fn algorithm(n: usize) void {\n    var a: i32 = 1; // +1\n    a += 1; // +1\n    a *= 2; // +1\n    // Loop n times\n    for (0..n) |_| { // +1 (execute i ++ every round)\n        std.debug.print(\"{}\\n\", .{0}); // +1\n    }\n}\n

Given a function that represents the number of operations of an algorithm as a function of the input size \\(n\\), denoted as \\(T(n)\\), consider the following example:

\\[ T(n) = 3 + 2n \\]

Since \\(T(n)\\) is a linear function, its growth trend is linear, and therefore, its time complexity is of linear order, denoted as \\(O(n)\\). This mathematical notation, known as \"big-O notation,\" represents the \"asymptotic upper bound\" of the function \\(T(n)\\).

In essence, time complexity analysis is about finding the asymptotic upper bound of the \"number of operations \\(T(n)\\)\". It has a precise mathematical definition.

Asymptotic Upper Bound

If there exist positive real numbers \\(c\\) and \\(n_0\\) such that for all \\(n > n_0\\), \\(T(n) \\leq c \\cdot f(n)\\), then \\(f(n)\\) is considered an asymptotic upper bound of \\(T(n)\\), denoted as \\(T(n) = O(f(n))\\).

As illustrated below, calculating the asymptotic upper bound involves finding a function \\(f(n)\\) such that, as \\(n\\) approaches infinity, \\(T(n)\\) and \\(f(n)\\) have the same growth order, differing only by a constant factor \\(c\\).

Figure 2-8 \u00a0 Asymptotic Upper Bound of a Function

"},{"location":"chapter_computational_complexity/time_complexity/#233-calculation-method","title":"2.3.3 \u00a0 Calculation Method","text":"

While the concept of asymptotic upper bound might seem mathematically dense, you don't need to fully grasp it right away. Let's first understand the method of calculation, which can be practiced and comprehended over time.

Once \\(f(n)\\) is determined, we obtain the time complexity \\(O(f(n))\\). But how do we determine the asymptotic upper bound \\(f(n)\\)? This process generally involves two steps: counting the number of operations and determining the asymptotic upper bound.

"},{"location":"chapter_computational_complexity/time_complexity/#1-step-1-counting-the-number-of-operations","title":"1. \u00a0 Step 1: Counting the Number of Operations","text":"

This step involves going through the code line by line. However, due to the presence of the constant \\(c\\) in \\(c \\cdot f(n)\\), all coefficients and constant terms in \\(T(n)\\) can be ignored. This principle allows for simplification techniques in counting operations.

  1. Ignore constant terms in \\(T(n)\\), as they do not affect the time complexity being independent of \\(n\\).
  2. Omit all coefficients. For example, looping \\(2n\\), \\(5n + 1\\) times, etc., can be simplified to \\(n\\) times since the coefficient before \\(n\\) does not impact the time complexity.
  3. Use multiplication for nested loops. The total number of operations equals the product of the number of operations in each loop, applying the simplification techniques from points 1 and 2 for each loop level.

Given a function, we can use these techniques to count operations:

PythonC++JavaC#GoSwiftJSTSDartRustCZig 
def algorithm(n: int):\n    a = 1      # +0 (trick 1)\n    a = a + n  # +0 (trick 1)\n    # +n (technique 2)\n    for i in range(5 * n + 1):\n        print(0)\n    # +n*n (technique 3)\n    for i in range(2 * n):\n        for j in range(n + 1):\n            print(0)\n
void algorithm(int n) {\n    int a = 1;  // +0 (trick 1)\n    a = a + n;  // +0 (trick 1)\n    // +n (technique 2)\n    for (int i = 0; i < 5 * n + 1; i++) {\n        cout << 0 << endl;\n    }\n    // +n*n (technique 3)\n    for (int i = 0; i < 2 * n; i++) {\n        for (int j = 0; j < n + 1; j++) {\n            cout << 0 << endl;\n        }\n    }\n}\n
void algorithm(int n) {\n    int a = 1;  // +0 (trick 1)\n    a = a + n;  // +0 (trick 1)\n    // +n (technique 2)\n    for (int i = 0; i < 5 * n + 1; i++) {\n        System.out.println(0);\n    }\n    // +n*n (technique 3)\n    for (int i = 0; i < 2 * n; i++) {\n        for (int j = 0; j < n + 1; j++) {\n            System.out.println(0);\n        }\n    }\n}\n
void Algorithm(int n) {\n    int a = 1;  // +0 (trick 1)\n    a = a + n;  // +0 (trick 1)\n    // +n (technique 2)\n    for (int i = 0; i < 5 * n + 1; i++) {\n        Console.WriteLine(0);\n    }\n    // +n*n (technique 3)\n    for (int i = 0; i < 2 * n; i++) {\n        for (int j = 0; j < n + 1; j++) {\n            Console.WriteLine(0);\n        }\n    }\n}\n
func algorithm(n int) {\n    a := 1     // +0 (trick 1)\n    a = a + n  // +0 (trick 1)\n    // +n (technique 2)\n    for i := 0; i < 5 * n + 1; i++ {\n        fmt.Println(0)\n    }\n    // +n*n (technique 3)\n    for i := 0; i < 2 * n; i++ {\n        for j := 0; j < n + 1; j++ {\n            fmt.Println(0)\n        }\n    }\n}\n
func algorithm(n: Int) {\n    var a = 1 // +0 (trick 1)\n    a = a + n // +0 (trick 1)\n    // +n (technique 2)\n    for _ in 0 ..< (5 * n + 1) {\n        print(0)\n    }\n    // +n*n (technique 3)\n    for _ in 0 ..< (2 * n) {\n        for _ in 0 ..< (n + 1) {\n            print(0)\n        }\n    }\n}\n
function algorithm(n) {\n    let a = 1;  // +0 (trick 1)\n    a = a + n;  // +0 (trick 1)\n    // +n (technique 2)\n    for (let i = 0; i < 5 * n + 1; i++) {\n        console.log(0);\n    }\n    // +n*n (technique 3)\n    for (let i = 0; i < 2 * n; i++) {\n        for (let j = 0; j < n + 1; j++) {\n            console.log(0);\n        }\n    }\n}\n
function algorithm(n: number): void {\n    let a = 1;  // +0 (trick 1)\n    a = a + n;  // +0 (trick 1)\n    // +n (technique 2)\n    for (let i = 0; i < 5 * n + 1; i++) {\n        console.log(0);\n    }\n    // +n*n (technique 3)\n    for (let i = 0; i < 2 * n; i++) {\n        for (let j = 0; j < n + 1; j++) {\n            console.log(0);\n        }\n    }\n}\n
void algorithm(int n) {\n  int a = 1; // +0 (trick 1)\n  a = a + n; // +0 (trick 1)\n  // +n (technique 2)\n  for (int i = 0; i < 5 * n + 1; i++) {\n    print(0);\n  }\n  // +n*n (technique 3)\n  for (int i = 0; i < 2 * n; i++) {\n    for (int j = 0; j < n + 1; j++) {\n      print(0);\n    }\n  }\n}\n
fn algorithm(n: i32) {\n    let mut a = 1;     // +0 (trick 1)\n    a = a + n;        // +0 (trick 1)\n\n    // +n (technique 2)\n    for i in 0..(5 * n + 1) {\n        println!(\"{}\", 0);\n    }\n\n    // +n*n (technique 3)\n    for i in 0..(2 * n) {\n        for j in 0..(n + 1) {\n            println!(\"{}\", 0);\n        }\n    }\n}\n
void algorithm(int n) {\n    int a = 1;  // +0 (trick 1)\n    a = a + n;  // +0 (trick 1)\n    // +n (technique 2)\n    for (int i = 0; i < 5 * n + 1; i++) {\n        printf(\"%d\", 0);\n    }\n    // +n*n (technique 3)\n    for (int i = 0; i < 2 * n; i++) {\n        for (int j = 0; j < n + 1; j++) {\n            printf(\"%d\", 0);\n        }\n    }\n}\n
fn algorithm(n: usize) void {\n    var a: i32 = 1;     // +0 (trick 1)\n    a = a + @as(i32, @intCast(n));        // +0 (trick 1)\n\n    // +n (technique 2)\n    for(0..(5 * n + 1)) |_| {\n        std.debug.print(\"{}\\n\", .{0});\n    }\n\n    // +n*n (technique 3)\n    for(0..(2 * n)) |_| {\n        for(0..(n + 1)) |_| {\n            std.debug.print(\"{}\\n\", .{0});\n        }\n    }\n}\n

The formula below shows the counting results before and after simplification, both leading to a time complexity of \\(O(n^2)\\):

\\[ \\begin{aligned} T(n) & = 2n(n + 1) + (5n + 1) + 2 & \\text{Complete Count (-.-|||)} \\newline & = 2n^2 + 7n + 3 \\newline T(n) & = n^2 + n & \\text{Simplified Count (o.O)} \\end{aligned} \\]"},{"location":"chapter_computational_complexity/time_complexity/#2-step-2-determining-the-asymptotic-upper-bound","title":"2. \u00a0 Step 2: Determining the Asymptotic Upper Bound","text":"

The time complexity is determined by the highest order term in \\(T(n)\\). This is because, as \\(n\\) approaches infinity, the highest order term dominates, rendering the influence of other terms negligible.

The following table illustrates examples of different operation counts and their corresponding time complexities. Some exaggerated values are used to emphasize that coefficients cannot alter the order of growth. When \\(n\\) becomes very large, these constants become insignificant.

Table: Time Complexity for Different Operation Counts

Operation Count \\(T(n)\\) Time Complexity \\(O(f(n))\\) \\(100000\\) \\(O(1)\\) \\(3n + 2\\) \\(O(n)\\) \\(2n^2 + 3n + 2\\) \\(O(n^2)\\) \\(n^3 + 10000n^2\\) \\(O(n^3)\\) \\(2^n + 10000n^{10000}\\) \\(O(2^n)\\)"},{"location":"chapter_computational_complexity/time_complexity/#234-common-types-of-time-complexity","title":"2.3.4 \u00a0 Common Types of Time Complexity","text":"

Let's consider the input data size as \\(n\\). The common types of time complexities are illustrated below, arranged from lowest to highest:

\\[ \\begin{aligned} O(1) < O(\\log n) < O(n) < O(n \\log n) < O(n^2) < O(2^n) < O(n!) \\newline \\text{Constant Order} < \\text{Logarithmic Order} < \\text{Linear Order} < \\text{Linear-Logarithmic Order} < \\text{Quadratic Order} < \\text{Exponential Order} < \\text{Factorial Order} \\end{aligned} \\]

Figure 2-9 \u00a0 Common Types of Time Complexity

"},{"location":"chapter_computational_complexity/time_complexity/#1-constant-order-o1","title":"1. \u00a0 Constant Order \\(O(1)\\)","text":"

Constant order means the number of operations is independent of the input data size \\(n\\). In the following function, although the number of operations size might be large, the time complexity remains \\(O(1)\\) as it's unrelated to \\(n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def constant(n: int) -> int:\n    \"\"\"\u5e38\u6570\u9636\"\"\"\n    count = 0\n    size = 100000\n    for _ in range(size):\n        count += 1\n    return count\n
time_complexity.cpp
/* \u5e38\u6570\u9636 */\nint constant(int n) {\n    int count = 0;\n    int size = 100000;\n    for (int i = 0; i < size; i++)\n        count++;\n    return count;\n}\n
time_complexity.java
/* \u5e38\u6570\u9636 */\nint constant(int n) {\n    int count = 0;\n    int size = 100000;\n    for (int i = 0; i < size; i++)\n        count++;\n    return count;\n}\n
time_complexity.cs
/* \u5e38\u6570\u9636 */\nint Constant(int n) {\n    int count = 0;\n    int size = 100000;\n    for (int i = 0; i < size; i++)\n        count++;\n    return count;\n}\n
time_complexity.go
/* \u5e38\u6570\u9636 */\nfunc constant(n int) int {\n    count := 0\n    size := 100000\n    for i := 0; i < size; i++ {\n        count++\n    }\n    return count\n}\n
time_complexity.swift
/* \u5e38\u6570\u9636 */\nfunc constant(n: Int) -> Int {\n    var count = 0\n    let size = 100_000\n    for _ in 0 ..< size {\n        count += 1\n    }\n    return count\n}\n
time_complexity.js
/* \u5e38\u6570\u9636 */\nfunction constant(n) {\n    let count = 0;\n    const size = 100000;\n    for (let i = 0; i < size; i++) count++;\n    return count;\n}\n
time_complexity.ts
/* \u5e38\u6570\u9636 */\nfunction constant(n: number): number {\n    let count = 0;\n    const size = 100000;\n    for (let i = 0; i < size; i++) count++;\n    return count;\n}\n
time_complexity.dart
/* \u5e38\u6570\u9636 */\nint constant(int n) {\n  int count = 0;\n  int size = 100000;\n  for (var i = 0; i < size; i++) {\n    count++;\n  }\n  return count;\n}\n
time_complexity.rs
/* \u5e38\u6570\u9636 */\nfn constant(n: i32) -> i32 {\n    _ = n;\n    let mut count = 0;\n    let size = 100_000;\n    for _ in 0..size {\n        count += 1;\n    }\n    count\n}\n
time_complexity.c
/* \u5e38\u6570\u9636 */\nint constant(int n) {\n    int count = 0;\n    int size = 100000;\n    int i = 0;\n    for (int i = 0; i < size; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.zig
// \u5e38\u6570\u9636\nfn constant(n: i32) i32 {\n    _ = n;\n    var count: i32 = 0;\n    const size: i32 = 100_000;\n    var i: i32 = 0;\n    while(i<size) : (i += 1) {\n        count += 1;\n    }\n    return count;\n}\n
"},{"location":"chapter_computational_complexity/time_complexity/#2-linear-order-on","title":"2. \u00a0 Linear Order \\(O(n)\\)","text":"

Linear order indicates the number of operations grows linearly with the input data size \\(n\\). Linear order commonly appears in single-loop structures:

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def linear(n: int) -> int:\n    \"\"\"\u7ebf\u6027\u9636\"\"\"\n    count = 0\n    for _ in range(n):\n        count += 1\n    return count\n
time_complexity.cpp
/* \u7ebf\u6027\u9636 */\nint linear(int n) {\n    int count = 0;\n    for (int i = 0; i < n; i++)\n        count++;\n    return count;\n}\n
time_complexity.java
/* \u7ebf\u6027\u9636 */\nint linear(int n) {\n    int count = 0;\n    for (int i = 0; i < n; i++)\n        count++;\n    return count;\n}\n
time_complexity.cs
/* \u7ebf\u6027\u9636 */\nint Linear(int n) {\n    int count = 0;\n    for (int i = 0; i < n; i++)\n        count++;\n    return count;\n}\n
time_complexity.go
/* \u7ebf\u6027\u9636 */\nfunc linear(n int) int {\n    count := 0\n    for i := 0; i < n; i++ {\n        count++\n    }\n    return count\n}\n
time_complexity.swift
/* \u7ebf\u6027\u9636 */\nfunc linear(n: Int) -> Int {\n    var count = 0\n    for _ in 0 ..< n {\n        count += 1\n    }\n    return count\n}\n
time_complexity.js
/* \u7ebf\u6027\u9636 */\nfunction linear(n) {\n    let count = 0;\n    for (let i = 0; i < n; i++) count++;\n    return count;\n}\n
time_complexity.ts
/* \u7ebf\u6027\u9636 */\nfunction linear(n: number): number {\n    let count = 0;\n    for (let i = 0; i < n; i++) count++;\n    return count;\n}\n
time_complexity.dart
/* \u7ebf\u6027\u9636 */\nint linear(int n) {\n  int count = 0;\n  for (var i = 0; i < n; i++) {\n    count++;\n  }\n  return count;\n}\n
time_complexity.rs
/* \u7ebf\u6027\u9636 */\nfn linear(n: i32) -> i32 {\n    let mut count = 0;\n    for _ in 0..n {\n        count += 1;\n    }\n    count\n}\n
time_complexity.c
/* \u7ebf\u6027\u9636 */\nint linear(int n) {\n    int count = 0;\n    for (int i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.zig
// \u7ebf\u6027\u9636\nfn linear(n: i32) i32 {\n    var count: i32 = 0;\n    var i: i32 = 0;\n    while (i < n) : (i += 1) {\n        count += 1;\n    }\n    return count;\n}\n

Operations like array traversal and linked list traversal have a time complexity of \\(O(n)\\), where \\(n\\) is the length of the array or list:

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def array_traversal(nums: list[int]) -> int:\n    \"\"\"\u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09\"\"\"\n    count = 0\n    # \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for num in nums:\n        count += 1\n    return count\n
time_complexity.cpp
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nint arrayTraversal(vector<int> &nums) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for (int num : nums) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.java
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nint arrayTraversal(int[] nums) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for (int num : nums) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.cs
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nint ArrayTraversal(int[] nums) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    foreach (int num in nums) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.go
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nfunc arrayTraversal(nums []int) int {\n    count := 0\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for range nums {\n        count++\n    }\n    return count\n}\n
time_complexity.swift
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nfunc arrayTraversal(nums: [Int]) -> Int {\n    var count = 0\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for _ in nums {\n        count += 1\n    }\n    return count\n}\n
time_complexity.js
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nfunction arrayTraversal(nums) {\n    let count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for (let i = 0; i < nums.length; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.ts
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nfunction arrayTraversal(nums: number[]): number {\n    let count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for (let i = 0; i < nums.length; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.dart
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nint arrayTraversal(List<int> nums) {\n  int count = 0;\n  // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n  for (var _num in nums) {\n    count++;\n  }\n  return count;\n}\n
time_complexity.rs
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nfn array_traversal(nums: &[i32]) -> i32 {\n    let mut count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for _ in nums {\n        count += 1;\n    }\n    count\n}\n
time_complexity.c
/* \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09 */\nint arrayTraversal(int *nums, int n) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for (int i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.zig
// \u7ebf\u6027\u9636\uff08\u904d\u5386\u6570\u7ec4\uff09\nfn arrayTraversal(nums: []i32) i32 {\n    var count: i32 = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u6b63\u6bd4\n    for (nums) |_| {\n        count += 1;\n    }\n    return count;\n}\n

It's important to note that the input data size \\(n\\) should be determined based on the type of input data. For example, in the first example, \\(n\\) represents the input data size, while in the second example, the length of the array \\(n\\) is the data size.

"},{"location":"chapter_computational_complexity/time_complexity/#3-quadratic-order-on2","title":"3. \u00a0 Quadratic Order \\(O(n^2)\\)","text":"

Quadratic order means the number of operations grows quadratically with the input data size \\(n\\). Quadratic order typically appears in nested loops, where both the outer and inner loops have a time complexity of \\(O(n)\\), resulting in an overall complexity of \\(O(n^2)\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def quadratic(n: int) -> int:\n    \"\"\"\u5e73\u65b9\u9636\"\"\"\n    count = 0\n    # \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for i in range(n):\n        for j in range(n):\n            count += 1\n    return count\n
time_complexity.cpp
/* \u5e73\u65b9\u9636 */\nint quadratic(int n) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < n; j++) {\n            count++;\n        }\n    }\n    return count;\n}\n
time_complexity.java
/* \u5e73\u65b9\u9636 */\nint quadratic(int n) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < n; j++) {\n            count++;\n        }\n    }\n    return count;\n}\n
time_complexity.cs
/* \u5e73\u65b9\u9636 */\nint Quadratic(int n) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < n; j++) {\n            count++;\n        }\n    }\n    return count;\n}\n
time_complexity.go
/* \u5e73\u65b9\u9636 */\nfunc quadratic(n int) int {\n    count := 0\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for i := 0; i < n; i++ {\n        for j := 0; j < n; j++ {\n            count++\n        }\n    }\n    return count\n}\n
time_complexity.swift
/* \u5e73\u65b9\u9636 */\nfunc quadratic(n: Int) -> Int {\n    var count = 0\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for _ in 0 ..< n {\n        for _ in 0 ..< n {\n            count += 1\n        }\n    }\n    return count\n}\n
time_complexity.js
/* \u5e73\u65b9\u9636 */\nfunction quadratic(n) {\n    let count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for (let i = 0; i < n; i++) {\n        for (let j = 0; j < n; j++) {\n            count++;\n        }\n    }\n    return count;\n}\n
time_complexity.ts
/* \u5e73\u65b9\u9636 */\nfunction quadratic(n: number): number {\n    let count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for (let i = 0; i < n; i++) {\n        for (let j = 0; j < n; j++) {\n            count++;\n        }\n    }\n    return count;\n}\n
time_complexity.dart
/* \u5e73\u65b9\u9636 */\nint quadratic(int n) {\n  int count = 0;\n  // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n  for (int i = 0; i < n; i++) {\n    for (int j = 0; j < n; j++) {\n      count++;\n    }\n  }\n  return count;\n}\n
time_complexity.rs
/* \u5e73\u65b9\u9636 */\nfn quadratic(n: i32) -> i32 {\n    let mut count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for _ in 0..n {\n        for _ in 0..n {\n            count += 1;\n        }\n    }\n    count\n}\n
time_complexity.c
/* \u5e73\u65b9\u9636 */\nint quadratic(int n) {\n    int count = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < n; j++) {\n            count++;\n        }\n    }\n    return count;\n}\n
time_complexity.zig
// \u5e73\u65b9\u9636\nfn quadratic(n: i32) i32 {\n    var count: i32 = 0;\n    var i: i32 = 0;\n    // \u5faa\u73af\u6b21\u6570\u4e0e\u6570\u7ec4\u957f\u5ea6\u6210\u5e73\u65b9\u5173\u7cfb\n    while (i < n) : (i += 1) {\n        var j: i32 = 0;\n        while (j < n) : (j += 1) {\n            count += 1;\n        }\n    }\n    return count;\n}\n

The following image compares constant order, linear order, and quadratic order time complexities.

Figure 2-10 \u00a0 Constant, Linear, and Quadratic Order Time Complexities

For instance, in bubble sort, the outer loop runs \\(n - 1\\) times, and the inner loop runs \\(n-1\\), \\(n-2\\), ..., \\(2\\), \\(1\\) times, averaging \\(n / 2\\) times, resulting in a time complexity of \\(O((n - 1) n / 2) = O(n^2)\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def bubble_sort(nums: list[int]) -> int:\n    \"\"\"\u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09\"\"\"\n    count = 0  # \u8ba1\u6570\u5668\n    # \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for i in range(len(nums) - 1, 0, -1):\n        # \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n        for j in range(i):\n            if nums[j] > nums[j + 1]:\n                # \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                tmp: int = nums[j]\n                nums[j] = nums[j + 1]\n                nums[j + 1] = tmp\n                count += 3  # \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n    return count\n
time_complexity.cpp
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nint bubbleSort(vector<int> &nums) {\n    int count = 0; // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for (int i = nums.size() - 1; i > 0; i--) {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n        for (int j = 0; j < i; j++) {\n            if (nums[j] > nums[j + 1]) {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                int tmp = nums[j];\n                nums[j] = nums[j + 1];\n                nums[j + 1] = tmp;\n                count += 3; // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count;\n}\n
time_complexity.java
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nint bubbleSort(int[] nums) {\n    int count = 0; // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for (int i = nums.length - 1; i > 0; i--) {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n        for (int j = 0; j < i; j++) {\n            if (nums[j] > nums[j + 1]) {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                int tmp = nums[j];\n                nums[j] = nums[j + 1];\n                nums[j + 1] = tmp;\n                count += 3; // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count;\n}\n
time_complexity.cs
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nint BubbleSort(int[] nums) {\n    int count = 0;  // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for (int i = nums.Length - 1; i > 0; i--) {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef \n        for (int j = 0; j < i; j++) {\n            if (nums[j] > nums[j + 1]) {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                (nums[j + 1], nums[j]) = (nums[j], nums[j + 1]);\n                count += 3;  // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count;\n}\n
time_complexity.go
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nfunc bubbleSort(nums []int) int {\n    count := 0 // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for i := len(nums) - 1; i > 0; i-- {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n        for j := 0; j < i; j++ {\n            if nums[j] > nums[j+1] {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                tmp := nums[j]\n                nums[j] = nums[j+1]\n                nums[j+1] = tmp\n                count += 3 // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count\n}\n
time_complexity.swift
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nfunc bubbleSort(nums: inout [Int]) -> Int {\n    var count = 0 // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for i in stride(from: nums.count - 1, to: 0, by: -1) {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef \n        for j in 0 ..< i {\n            if nums[j] > nums[j + 1] {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                let tmp = nums[j]\n                nums[j] = nums[j + 1]\n                nums[j + 1] = tmp\n                count += 3 // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count\n}\n
time_complexity.js
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nfunction bubbleSort(nums) {\n    let count = 0; // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for (let i = nums.length - 1; i > 0; i--) {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n        for (let j = 0; j < i; j++) {\n            if (nums[j] > nums[j + 1]) {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                let tmp = nums[j];\n                nums[j] = nums[j + 1];\n                nums[j + 1] = tmp;\n                count += 3; // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count;\n}\n
time_complexity.ts
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nfunction bubbleSort(nums: number[]): number {\n    let count = 0; // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for (let i = nums.length - 1; i > 0; i--) {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n        for (let j = 0; j < i; j++) {\n            if (nums[j] > nums[j + 1]) {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                let tmp = nums[j];\n                nums[j] = nums[j + 1];\n                nums[j + 1] = tmp;\n                count += 3; // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count;\n}\n
time_complexity.dart
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nint bubbleSort(List<int> nums) {\n  int count = 0; // \u8ba1\u6570\u5668\n  // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n  for (var i = nums.length - 1; i > 0; i--) {\n    // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n    for (var j = 0; j < i; j++) {\n      if (nums[j] > nums[j + 1]) {\n        // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n        int tmp = nums[j];\n        nums[j] = nums[j + 1];\n        nums[j + 1] = tmp;\n        count += 3; // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n      }\n    }\n  }\n  return count;\n}\n
time_complexity.rs
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nfn bubble_sort(nums: &mut [i32]) -> i32 {\n    let mut count = 0; // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for i in (1..nums.len()).rev() {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef \n        for j in 0..i {\n            if nums[j] > nums[j + 1] {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                let tmp = nums[j];\n                nums[j] = nums[j + 1];\n                nums[j + 1] = tmp;\n                count += 3; // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    count\n}\n
time_complexity.c
/* \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09 */\nint bubbleSort(int *nums, int n) {\n    int count = 0; // \u8ba1\u6570\u5668\n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    for (int i = n - 1; i > 0; i--) {\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef\n        for (int j = 0; j < i; j++) {\n            if (nums[j] > nums[j + 1]) {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                int tmp = nums[j];\n                nums[j] = nums[j + 1];\n                nums[j + 1] = tmp;\n                count += 3; // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count;\n}\n
time_complexity.zig
// \u5e73\u65b9\u9636\uff08\u5192\u6ce1\u6392\u5e8f\uff09\nfn bubbleSort(nums: []i32) i32 {\n    var count: i32 = 0;  // \u8ba1\u6570\u5668 \n    // \u5916\u5faa\u73af\uff1a\u672a\u6392\u5e8f\u533a\u95f4\u4e3a [0, i]\n    var i: i32 = @as(i32, @intCast(nums.len)) - 1;\n    while (i > 0) : (i -= 1) {\n        var j: usize = 0;\n        // \u5185\u5faa\u73af\uff1a\u5c06\u672a\u6392\u5e8f\u533a\u95f4 [0, i] \u4e2d\u7684\u6700\u5927\u5143\u7d20\u4ea4\u6362\u81f3\u8be5\u533a\u95f4\u7684\u6700\u53f3\u7aef \n        while (j < i) : (j += 1) {\n            if (nums[j] > nums[j + 1]) {\n                // \u4ea4\u6362 nums[j] \u4e0e nums[j + 1]\n                var tmp = nums[j];\n                nums[j] = nums[j + 1];\n                nums[j + 1] = tmp;\n                count += 3;  // \u5143\u7d20\u4ea4\u6362\u5305\u542b 3 \u4e2a\u5355\u5143\u64cd\u4f5c\n            }\n        }\n    }\n    return count;\n}\n
"},{"location":"chapter_computational_complexity/time_complexity/#4-exponential-order-o2n","title":"4. \u00a0 Exponential Order \\(O(2^n)\\)","text":"

Biological \"cell division\" is a classic example of exponential order growth: starting with one cell, it becomes two after one division, four after two divisions, and so on, resulting in \\(2^n\\) cells after \\(n\\) divisions.

The following image and code simulate the cell division process, with a time complexity of \\(O(2^n)\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def exponential(n: int) -> int:\n    \"\"\"\u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09\"\"\"\n    count = 0\n    base = 1\n    # \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for _ in range(n):\n        for _ in range(base):\n            count += 1\n        base *= 2\n    # count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count\n
time_complexity.cpp
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint exponential(int n) {\n    int count = 0, base = 1;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < base; j++) {\n            count++;\n        }\n        base *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count;\n}\n
time_complexity.java
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint exponential(int n) {\n    int count = 0, base = 1;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < base; j++) {\n            count++;\n        }\n        base *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count;\n}\n
time_complexity.cs
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint Exponential(int n) {\n    int count = 0, bas = 1;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < bas; j++) {\n            count++;\n        }\n        bas *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count;\n}\n
time_complexity.go
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09*/\nfunc exponential(n int) int {\n    count, base := 0, 1\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for i := 0; i < n; i++ {\n        for j := 0; j < base; j++ {\n            count++\n        }\n        base *= 2\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count\n}\n
time_complexity.swift
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfunc exponential(n: Int) -> Int {\n    var count = 0\n    var base = 1\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for _ in 0 ..< n {\n        for _ in 0 ..< base {\n            count += 1\n        }\n        base *= 2\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count\n}\n
time_complexity.js
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfunction exponential(n) {\n    let count = 0,\n        base = 1;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for (let i = 0; i < n; i++) {\n        for (let j = 0; j < base; j++) {\n            count++;\n        }\n        base *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count;\n}\n
time_complexity.ts
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfunction exponential(n: number): number {\n    let count = 0,\n        base = 1;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for (let i = 0; i < n; i++) {\n        for (let j = 0; j < base; j++) {\n            count++;\n        }\n        base *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count;\n}\n
time_complexity.dart
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint exponential(int n) {\n  int count = 0, base = 1;\n  // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n  for (var i = 0; i < n; i++) {\n    for (var j = 0; j < base; j++) {\n      count++;\n    }\n    base *= 2;\n  }\n  // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n  return count;\n}\n
time_complexity.rs
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfn exponential(n: i32) -> i32 {\n    let mut count = 0;\n    let mut base = 1;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for _ in 0..n {\n        for _ in 0..base {\n            count += 1\n        }\n        base *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    count\n}\n
time_complexity.c
/* \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint exponential(int n) {\n    int count = 0;\n    int bas = 1;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    for (int i = 0; i < n; i++) {\n        for (int j = 0; j < bas; j++) {\n            count++;\n        }\n        bas *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count;\n}\n
time_complexity.zig
// \u6307\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09\nfn exponential(n: i32) i32 {\n    var count: i32 = 0;\n    var bas: i32 = 1;\n    var i: i32 = 0;\n    // \u7ec6\u80de\u6bcf\u8f6e\u4e00\u5206\u4e3a\u4e8c\uff0c\u5f62\u6210\u6570\u5217 1, 2, 4, 8, ..., 2^(n-1)\n    while (i < n) : (i += 1) {\n        var j: i32 = 0;\n        while (j < bas) : (j += 1) {\n            count += 1;\n        }\n        bas *= 2;\n    }\n    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1\n    return count;\n}\n

Figure 2-11 \u00a0 Exponential Order Time Complexity

In practice, exponential order often appears in recursive functions. For example, in the code below, it recursively splits into two halves, stopping after \\(n\\) divisions:

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def exp_recur(n: int) -> int:\n    \"\"\"\u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\"\"\"\n    if n == 1:\n        return 1\n    return exp_recur(n - 1) + exp_recur(n - 1) + 1\n
time_complexity.cpp
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint expRecur(int n) {\n    if (n == 1)\n        return 1;\n    return expRecur(n - 1) + expRecur(n - 1) + 1;\n}\n
time_complexity.java
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint expRecur(int n) {\n    if (n == 1)\n        return 1;\n    return expRecur(n - 1) + expRecur(n - 1) + 1;\n}\n
time_complexity.cs
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint ExpRecur(int n) {\n    if (n == 1) return 1;\n    return ExpRecur(n - 1) + ExpRecur(n - 1) + 1;\n}\n
time_complexity.go
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09*/\nfunc expRecur(n int) int {\n    if n == 1 {\n        return 1\n    }\n    return expRecur(n-1) + expRecur(n-1) + 1\n}\n
time_complexity.swift
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunc expRecur(n: Int) -> Int {\n    if n == 1 {\n        return 1\n    }\n    return expRecur(n: n - 1) + expRecur(n: n - 1) + 1\n}\n
time_complexity.js
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction expRecur(n) {\n    if (n === 1) return 1;\n    return expRecur(n - 1) + expRecur(n - 1) + 1;\n}\n
time_complexity.ts
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction expRecur(n: number): number {\n    if (n === 1) return 1;\n    return expRecur(n - 1) + expRecur(n - 1) + 1;\n}\n
time_complexity.dart
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint expRecur(int n) {\n  if (n == 1) return 1;\n  return expRecur(n - 1) + expRecur(n - 1) + 1;\n}\n
time_complexity.rs
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfn exp_recur(n: i32) -> i32 {\n    if n == 1 {\n        return 1;\n    }\n    exp_recur(n - 1) + exp_recur(n - 1) + 1\n}\n
time_complexity.c
/* \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint expRecur(int n) {\n    if (n == 1)\n        return 1;\n    return expRecur(n - 1) + expRecur(n - 1) + 1;\n}\n
time_complexity.zig
// \u6307\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\nfn expRecur(n: i32) i32 {\n    if (n == 1) return 1;\n    return expRecur(n - 1) + expRecur(n - 1) + 1;\n}\n

Exponential order growth is extremely rapid and is commonly seen in exhaustive search methods (brute force, backtracking, etc.). For large-scale problems, exponential order is unacceptable, often requiring dynamic programming or greedy algorithms as solutions.

"},{"location":"chapter_computational_complexity/time_complexity/#5-logarithmic-order-olog-n","title":"5. \u00a0 Logarithmic Order \\(O(\\log n)\\)","text":"

In contrast to exponential order, logarithmic order reflects situations where \"the size is halved each round.\" Given an input data size \\(n\\), since the size is halved each round, the number of iterations is \\(\\log_2 n\\), the inverse function of \\(2^n\\).

The following image and code simulate the \"halving each round\" process, with a time complexity of \\(O(\\log_2 n)\\), commonly abbreviated as \\(O(\\log n)\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def logarithmic(n: float) -> int:\n    \"\"\"\u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09\"\"\"\n    count = 0\n    while n > 1:\n        n = n / 2\n        count += 1\n    return count\n
time_complexity.cpp
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint logarithmic(float n) {\n    int count = 0;\n    while (n > 1) {\n        n = n / 2;\n        count++;\n    }\n    return count;\n}\n
time_complexity.java
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint logarithmic(float n) {\n    int count = 0;\n    while (n > 1) {\n        n = n / 2;\n        count++;\n    }\n    return count;\n}\n
time_complexity.cs
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint Logarithmic(float n) {\n    int count = 0;\n    while (n > 1) {\n        n /= 2;\n        count++;\n    }\n    return count;\n}\n
time_complexity.go
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09*/\nfunc logarithmic(n float64) int {\n    count := 0\n    for n > 1 {\n        n = n / 2\n        count++\n    }\n    return count\n}\n
time_complexity.swift
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfunc logarithmic(n: Double) -> Int {\n    var count = 0\n    var n = n\n    while n > 1 {\n        n = n / 2\n        count += 1\n    }\n    return count\n}\n
time_complexity.js
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfunction logarithmic(n) {\n    let count = 0;\n    while (n > 1) {\n        n = n / 2;\n        count++;\n    }\n    return count;\n}\n
time_complexity.ts
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfunction logarithmic(n: number): number {\n    let count = 0;\n    while (n > 1) {\n        n = n / 2;\n        count++;\n    }\n    return count;\n}\n
time_complexity.dart
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint logarithmic(num n) {\n  int count = 0;\n  while (n > 1) {\n    n = n / 2;\n    count++;\n  }\n  return count;\n}\n
time_complexity.rs
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nfn logarithmic(mut n: f32) -> i32 {\n    let mut count = 0;\n    while n > 1.0 {\n        n = n / 2.0;\n        count += 1;\n    }\n    count\n}\n
time_complexity.c
/* \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09 */\nint logarithmic(float n) {\n    int count = 0;\n    while (n > 1) {\n        n = n / 2;\n        count++;\n    }\n    return count;\n}\n
time_complexity.zig
// \u5bf9\u6570\u9636\uff08\u5faa\u73af\u5b9e\u73b0\uff09\nfn logarithmic(n: f32) i32 {\n    var count: i32 = 0;\n    var n_var = n;\n    while (n_var > 1)\n    {\n        n_var = n_var / 2;\n        count +=1;\n    }\n    return count;\n}\n

Figure 2-12 \u00a0 Logarithmic Order Time Complexity

Like exponential order, logarithmic order also frequently appears in recursive functions. The code below forms a recursive tree of height \\(\\log_2 n\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def log_recur(n: float) -> int:\n    \"\"\"\u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\"\"\"\n    if n <= 1:\n        return 0\n    return log_recur(n / 2) + 1\n
time_complexity.cpp
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint logRecur(float n) {\n    if (n <= 1)\n        return 0;\n    return logRecur(n / 2) + 1;\n}\n
time_complexity.java
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint logRecur(float n) {\n    if (n <= 1)\n        return 0;\n    return logRecur(n / 2) + 1;\n}\n
time_complexity.cs
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint LogRecur(float n) {\n    if (n <= 1) return 0;\n    return LogRecur(n / 2) + 1;\n}\n
time_complexity.go
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09*/\nfunc logRecur(n float64) int {\n    if n <= 1 {\n        return 0\n    }\n    return logRecur(n/2) + 1\n}\n
time_complexity.swift
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunc logRecur(n: Double) -> Int {\n    if n <= 1 {\n        return 0\n    }\n    return logRecur(n: n / 2) + 1\n}\n
time_complexity.js
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction logRecur(n) {\n    if (n <= 1) return 0;\n    return logRecur(n / 2) + 1;\n}\n
time_complexity.ts
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction logRecur(n: number): number {\n    if (n <= 1) return 0;\n    return logRecur(n / 2) + 1;\n}\n
time_complexity.dart
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint logRecur(num n) {\n  if (n <= 1) return 0;\n  return logRecur(n / 2) + 1;\n}\n
time_complexity.rs
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfn log_recur(n: f32) -> i32 {\n    if n <= 1.0 {\n        return 0;\n    }\n    log_recur(n / 2.0) + 1\n}\n
time_complexity.c
/* \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint logRecur(float n) {\n    if (n <= 1)\n        return 0;\n    return logRecur(n / 2) + 1;\n}\n
time_complexity.zig
// \u5bf9\u6570\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\nfn logRecur(n: f32) i32 {\n    if (n <= 1) return 0;\n    return logRecur(n / 2) + 1;\n}\n

Logarithmic order is typical in algorithms based on the divide-and-conquer strategy, embodying the \"split into many\" and \"simplify complex problems\" approach. It's slow-growing and is the most ideal time complexity after constant order.

What is the base of \\(O(\\log n)\\)?

Technically, \"splitting into \\(m\\)\" corresponds to a time complexity of \\(O(\\log_m n)\\). Using the logarithm base change formula, we can equate different logarithmic complexities:

\\[ O(\\log_m n) = O(\\log_k n / \\log_k m) = O(\\log_k n) \\]

This means the base \\(m\\) can be changed without affecting the complexity. Therefore, we often omit the base \\(m\\) and simply denote logarithmic order as \\(O(\\log n)\\).

"},{"location":"chapter_computational_complexity/time_complexity/#6-linear-logarithmic-order-on-log-n","title":"6. \u00a0 Linear-Logarithmic Order \\(O(n \\log n)\\)","text":"

Linear-logarithmic order often appears in nested loops, with the complexities of the two loops being \\(O(\\log n)\\) and \\(O(n)\\) respectively. The related code is as follows:

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def linear_log_recur(n: float) -> int:\n    \"\"\"\u7ebf\u6027\u5bf9\u6570\u9636\"\"\"\n    if n <= 1:\n        return 1\n    count: int = linear_log_recur(n // 2) + linear_log_recur(n // 2)\n    for _ in range(n):\n        count += 1\n    return count\n
time_complexity.cpp
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nint linearLogRecur(float n) {\n    if (n <= 1)\n        return 1;\n    int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);\n    for (int i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.java
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nint linearLogRecur(float n) {\n    if (n <= 1)\n        return 1;\n    int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);\n    for (int i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.cs
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nint LinearLogRecur(float n) {\n    if (n <= 1) return 1;\n    int count = LinearLogRecur(n / 2) + LinearLogRecur(n / 2);\n    for (int i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.go
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nfunc linearLogRecur(n float64) int {\n    if n <= 1 {\n        return 1\n    }\n    count := linearLogRecur(n/2) + linearLogRecur(n/2)\n    for i := 0.0; i < n; i++ {\n        count++\n    }\n    return count\n}\n
time_complexity.swift
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nfunc linearLogRecur(n: Double) -> Int {\n    if n <= 1 {\n        return 1\n    }\n    var count = linearLogRecur(n: n / 2) + linearLogRecur(n: n / 2)\n    for _ in stride(from: 0, to: n, by: 1) {\n        count += 1\n    }\n    return count\n}\n
time_complexity.js
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nfunction linearLogRecur(n) {\n    if (n <= 1) return 1;\n    let count = linearLogRecur(n / 2) + linearLogRecur(n / 2);\n    for (let i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.ts
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nfunction linearLogRecur(n: number): number {\n    if (n <= 1) return 1;\n    let count = linearLogRecur(n / 2) + linearLogRecur(n / 2);\n    for (let i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.dart
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nint linearLogRecur(num n) {\n  if (n <= 1) return 1;\n  int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);\n  for (var i = 0; i < n; i++) {\n    count++;\n  }\n  return count;\n}\n
time_complexity.rs
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nfn linear_log_recur(n: f32) -> i32 {\n    if n <= 1.0 {\n        return 1;\n    }\n    let mut count = linear_log_recur(n / 2.0) + linear_log_recur(n / 2.0);\n    for _ in 0 ..n as i32 {\n        count += 1;\n    }\n    return count\n}\n
time_complexity.c
/* \u7ebf\u6027\u5bf9\u6570\u9636 */\nint linearLogRecur(float n) {\n    if (n <= 1)\n        return 1;\n    int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);\n    for (int i = 0; i < n; i++) {\n        count++;\n    }\n    return count;\n}\n
time_complexity.zig
// \u7ebf\u6027\u5bf9\u6570\u9636\nfn linearLogRecur(n: f32) i32 {\n    if (n <= 1) return 1;\n    var count: i32 = linearLogRecur(n / 2) + linearLogRecur(n / 2);\n    var i: f32 = 0;\n    while (i < n) : (i += 1) {\n        count += 1;\n    }\n    return count;\n}\n

The image below demonstrates how linear-logarithmic order is generated. Each level of a binary tree has \\(n\\) operations, and the tree has \\(\\log_2 n + 1\\) levels, resulting in a time complexity of \\(O(n \\log n)\\).

Figure 2-13 \u00a0 Linear-Logarithmic Order Time Complexity

Mainstream sorting algorithms typically have a time complexity of \\(O(n \\log n)\\), such as quicksort, mergesort, and heapsort.

"},{"location":"chapter_computational_complexity/time_complexity/#7-factorial-order-on","title":"7. \u00a0 Factorial Order \\(O(n!)\\)","text":"

Factorial order corresponds to the mathematical problem of \"full permutation.\" Given \\(n\\) distinct elements, the total number of possible permutations is:

\\[ n! = n \\times (n - 1) \\times (n - 2) \\times \\dots \\times 2 \\times 1 \\]

Factorials are typically implemented using recursion. As shown in the image and code below, the first level splits into \\(n\\) branches, the second level into \\(n - 1\\) branches, and so on, stopping after the \\(n\\)th level:

PythonC++JavaC#GoSwiftJSTSDartRustCZig time_complexity.py
def factorial_recur(n: int) -> int:\n    \"\"\"\u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\"\"\"\n    if n == 0:\n        return 1\n    count = 0\n    # \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for _ in range(n):\n        count += factorial_recur(n - 1)\n    return count\n
time_complexity.cpp
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint factorialRecur(int n) {\n    if (n == 0)\n        return 1;\n    int count = 0;\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for (int i = 0; i < n; i++) {\n        count += factorialRecur(n - 1);\n    }\n    return count;\n}\n
time_complexity.java
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint factorialRecur(int n) {\n    if (n == 0)\n        return 1;\n    int count = 0;\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for (int i = 0; i < n; i++) {\n        count += factorialRecur(n - 1);\n    }\n    return count;\n}\n
time_complexity.cs
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint FactorialRecur(int n) {\n    if (n == 0) return 1;\n    int count = 0;\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for (int i = 0; i < n; i++) {\n        count += FactorialRecur(n - 1);\n    }\n    return count;\n}\n
time_complexity.go
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunc factorialRecur(n int) int {\n    if n == 0 {\n        return 1\n    }\n    count := 0\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for i := 0; i < n; i++ {\n        count += factorialRecur(n - 1)\n    }\n    return count\n}\n
time_complexity.swift
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunc factorialRecur(n: Int) -> Int {\n    if n == 0 {\n        return 1\n    }\n    var count = 0\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for _ in 0 ..< n {\n        count += factorialRecur(n: n - 1)\n    }\n    return count\n}\n
time_complexity.js
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction factorialRecur(n) {\n    if (n === 0) return 1;\n    let count = 0;\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for (let i = 0; i < n; i++) {\n        count += factorialRecur(n - 1);\n    }\n    return count;\n}\n
time_complexity.ts
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfunction factorialRecur(n: number): number {\n    if (n === 0) return 1;\n    let count = 0;\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for (let i = 0; i < n; i++) {\n        count += factorialRecur(n - 1);\n    }\n    return count;\n}\n
time_complexity.dart
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint factorialRecur(int n) {\n  if (n == 0) return 1;\n  int count = 0;\n  // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n  for (var i = 0; i < n; i++) {\n    count += factorialRecur(n - 1);\n  }\n  return count;\n}\n
time_complexity.rs
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nfn factorial_recur(n: i32) -> i32 {\n    if n == 0 {\n        return 1;\n    }\n    let mut count = 0;\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    for _ in 0..n {\n        count += factorial_recur(n - 1);\n    }\n    count\n}\n
time_complexity.c
/* \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09 */\nint factorialRecur(int n) {\n    if (n == 0)\n        return 1;\n    int count = 0;\n    for (int i = 0; i < n; i++) {\n        count += factorialRecur(n - 1);\n    }\n    return count;\n}\n
time_complexity.zig
// \u9636\u4e58\u9636\uff08\u9012\u5f52\u5b9e\u73b0\uff09\nfn factorialRecur(n: i32) i32 {\n    if (n == 0) return 1;\n    var count: i32 = 0;\n    var i: i32 = 0;\n    // \u4ece 1 \u4e2a\u5206\u88c2\u51fa n \u4e2a\n    while (i < n) : (i += 1) {\n        count += factorialRecur(n - 1);\n    }\n    return count;\n}\n

Figure 2-14 \u00a0 Factorial Order Time Complexity

Note that factorial order grows even faster than exponential order; it's unacceptable for larger \\(n\\) values.

"},{"location":"chapter_computational_complexity/time_complexity/#235-worst-best-and-average-time-complexities","title":"2.3.5 \u00a0 Worst, Best, and Average Time Complexities","text":"

The time efficiency of an algorithm is often not fixed but depends on the distribution of the input data. Assume we have an array nums of length \\(n\\), consisting of numbers from \\(1\\) to \\(n\\), each appearing only once, but in a randomly shuffled order. The task is to return the index of the element \\(1\\). We can draw the following conclusions:

The \"worst-case time complexity\" corresponds to the asymptotic upper bound, denoted by the big \\(O\\) notation. Correspondingly, the \"best-case time complexity\" corresponds to the asymptotic lower bound, denoted by \\(\\Omega\\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig worst_best_time_complexity.py
def random_numbers(n: int) -> list[int]:\n    \"\"\"\u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a: 1, 2, ..., n \uff0c\u987a\u5e8f\u88ab\u6253\u4e71\"\"\"\n    # \u751f\u6210\u6570\u7ec4 nums =: 1, 2, 3, ..., n\n    nums = [i for i in range(1, n + 1)]\n    # \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    random.shuffle(nums)\n    return nums\n\ndef find_one(nums: list[int]) -> int:\n    \"\"\"\u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15\"\"\"\n    for i in range(len(nums)):\n        # \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        # \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if nums[i] == 1:\n            return i\n    return -1\n
worst_best_time_complexity.cpp
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nvector<int> randomNumbers(int n) {\n    vector<int> nums(n);\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for (int i = 0; i < n; i++) {\n        nums[i] = i + 1;\n    }\n    // \u4f7f\u7528\u7cfb\u7edf\u65f6\u95f4\u751f\u6210\u968f\u673a\u79cd\u5b50\n    unsigned seed = chrono::system_clock::now().time_since_epoch().count();\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    shuffle(nums.begin(), nums.end(), default_random_engine(seed));\n    return nums;\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nint findOne(vector<int> &nums) {\n    for (int i = 0; i < nums.size(); i++) {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if (nums[i] == 1)\n            return i;\n    }\n    return -1;\n}\n
worst_best_time_complexity.java
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nint[] randomNumbers(int n) {\n    Integer[] nums = new Integer[n];\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for (int i = 0; i < n; i++) {\n        nums[i] = i + 1;\n    }\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    Collections.shuffle(Arrays.asList(nums));\n    // Integer[] -> int[]\n    int[] res = new int[n];\n    for (int i = 0; i < n; i++) {\n        res[i] = nums[i];\n    }\n    return res;\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nint findOne(int[] nums) {\n    for (int i = 0; i < nums.length; i++) {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if (nums[i] == 1)\n            return i;\n    }\n    return -1;\n}\n
worst_best_time_complexity.cs
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nint[] RandomNumbers(int n) {\n    int[] nums = new int[n];\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for (int i = 0; i < n; i++) {\n        nums[i] = i + 1;\n    }\n\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    for (int i = 0; i < nums.Length; i++) {\n        int index = new Random().Next(i, nums.Length);\n        (nums[i], nums[index]) = (nums[index], nums[i]);\n    }\n    return nums;\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nint FindOne(int[] nums) {\n    for (int i = 0; i < nums.Length; i++) {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if (nums[i] == 1)\n            return i;\n    }\n    return -1;\n}\n
worst_best_time_complexity.go
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nfunc randomNumbers(n int) []int {\n    nums := make([]int, n)\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for i := 0; i < n; i++ {\n        nums[i] = i + 1\n    }\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    rand.Shuffle(len(nums), func(i, j int) {\n        nums[i], nums[j] = nums[j], nums[i]\n    })\n    return nums\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nfunc findOne(nums []int) int {\n    for i := 0; i < len(nums); i++ {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if nums[i] == 1 {\n            return i\n        }\n    }\n    return -1\n}\n
worst_best_time_complexity.swift
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nfunc randomNumbers(n: Int) -> [Int] {\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    var nums = Array(1 ... n)\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    nums.shuffle()\n    return nums\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nfunc findOne(nums: [Int]) -> Int {\n    for i in nums.indices {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if nums[i] == 1 {\n            return i\n        }\n    }\n    return -1\n}\n
worst_best_time_complexity.js
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nfunction randomNumbers(n) {\n    const nums = Array(n);\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for (let i = 0; i < n; i++) {\n        nums[i] = i + 1;\n    }\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    for (let i = 0; i < n; i++) {\n        const r = Math.floor(Math.random() * (i + 1));\n        const temp = nums[i];\n        nums[i] = nums[r];\n        nums[r] = temp;\n    }\n    return nums;\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nfunction findOne(nums) {\n    for (let i = 0; i < nums.length; i++) {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if (nums[i] === 1) {\n            return i;\n        }\n    }\n    return -1;\n}\n
worst_best_time_complexity.ts
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nfunction randomNumbers(n: number): number[] {\n    const nums = Array(n);\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for (let i = 0; i < n; i++) {\n        nums[i] = i + 1;\n    }\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    for (let i = 0; i < n; i++) {\n        const r = Math.floor(Math.random() * (i + 1));\n        const temp = nums[i];\n        nums[i] = nums[r];\n        nums[r] = temp;\n    }\n    return nums;\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nfunction findOne(nums: number[]): number {\n    for (let i = 0; i < nums.length; i++) {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if (nums[i] === 1) {\n            return i;\n        }\n    }\n    return -1;\n}\n
worst_best_time_complexity.dart
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nList<int> randomNumbers(int n) {\n  final nums = List.filled(n, 0);\n  // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n  for (var i = 0; i < n; i++) {\n    nums[i] = i + 1;\n  }\n  // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n  nums.shuffle();\n\n  return nums;\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nint findOne(List<int> nums) {\n  for (var i = 0; i < nums.length; i++) {\n    // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n    // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n    if (nums[i] == 1) return i;\n  }\n\n  return -1;\n}\n
worst_best_time_complexity.rs
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nfn random_numbers(n: i32) -> Vec<i32> {\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    let mut nums = (1..=n).collect::<Vec<i32>>();\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    nums.shuffle(&mut thread_rng());\n    nums\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nfn find_one(nums: &[i32]) -> Option<usize> {\n    for i in 0..nums.len() {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if nums[i] == 1 {\n            return Some(i);\n        }\n    }\n    None\n}\n
worst_best_time_complexity.c
/* \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71 */\nint *randomNumbers(int n) {\n    // \u5206\u914d\u5806\u533a\u5185\u5b58\uff08\u521b\u5efa\u4e00\u7ef4\u53ef\u53d8\u957f\u6570\u7ec4\uff1a\u6570\u7ec4\u4e2d\u5143\u7d20\u6570\u91cf\u4e3a n \uff0c\u5143\u7d20\u7c7b\u578b\u4e3a int \uff09\n    int *nums = (int *)malloc(n * sizeof(int));\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for (int i = 0; i < n; i++) {\n        nums[i] = i + 1;\n    }\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    for (int i = n - 1; i > 0; i--) {\n        int j = rand() % (i + 1);\n        int temp = nums[i];\n        nums[i] = nums[j];\n        nums[j] = temp;\n    }\n    return nums;\n}\n\n/* \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15 */\nint findOne(int *nums, int n) {\n    for (int i = 0; i < n; i++) {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if (nums[i] == 1)\n            return i;\n    }\n    return -1;\n}\n
worst_best_time_complexity.zig
// \u751f\u6210\u4e00\u4e2a\u6570\u7ec4\uff0c\u5143\u7d20\u4e3a { 1, 2, ..., n }\uff0c\u987a\u5e8f\u88ab\u6253\u4e71\nfn randomNumbers(comptime n: usize) [n]i32 {\n    var nums: [n]i32 = undefined;\n    // \u751f\u6210\u6570\u7ec4 nums = { 1, 2, 3, ..., n }\n    for (&nums, 0..) |*num, i| {\n        num.* = @as(i32, @intCast(i)) + 1;\n    }\n    // \u968f\u673a\u6253\u4e71\u6570\u7ec4\u5143\u7d20\n    const rand = std.crypto.random;\n    rand.shuffle(i32, &nums);\n    return nums;\n}\n\n// \u67e5\u627e\u6570\u7ec4 nums \u4e2d\u6570\u5b57 1 \u6240\u5728\u7d22\u5f15\nfn findOne(nums: []i32) i32 {\n    for (nums, 0..) |num, i| {\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5934\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u4f73\u65f6\u95f4\u590d\u6742\u5ea6 O(1)\n        // \u5f53\u5143\u7d20 1 \u5728\u6570\u7ec4\u5c3e\u90e8\u65f6\uff0c\u8fbe\u5230\u6700\u5dee\u65f6\u95f4\u590d\u6742\u5ea6 O(n)\n        if (num == 1) return @intCast(i);\n    }\n    return -1;\n}\n

It's important to note that the best-case time complexity is rarely used in practice, as it is usually only achievable under very low probabilities and might be misleading. The worst-case time complexity is more practical as it provides a safety value for efficiency, allowing us to confidently use the algorithm.

From the above example, it's clear that both the worst-case and best-case time complexities only occur under \"special data distributions,\" which may have a small probability of occurrence and may not accurately reflect the algorithm's run efficiency. In contrast, the average time complexity can reflect the algorithm's efficiency under random input data, denoted by the \\(\\Theta\\) notation.

For some algorithms, we can simply estimate the average case under a random data distribution. For example, in the aforementioned example, since the input array is shuffled, the probability of element \\(1\\) appearing at any index is equal. Therefore, the average number of loops for the algorithm is half the length of the array \\(n / 2\\), giving an average time complexity of \\(\\Theta(n / 2) = \\Theta(n)\\).

However, calculating the average time complexity for more complex algorithms can be quite difficult, as it's challenging to analyze the overall mathematical expectation under the data distribution. In such cases, we usually use the worst-case time complexity as the standard for judging the efficiency of the algorithm.

Why is the \\(\\Theta\\) symbol rarely seen?

Possibly because the \\(O\\) notation is more commonly spoken, it is often used to represent the average time complexity. However, strictly speaking, this practice is not accurate. In this book and other materials, if you encounter statements like \"average time complexity \\(O(n)\\)\", please understand it directly as \\(\\Theta(n)\\).

"},{"location":"chapter_data_structure/","title":"Data Structure","text":"

Abstract

Data structures resemble a stable and diverse framework.

They serve as a blueprint for organizing data orderly, enabling algorithms to come to life upon this foundation.

"},{"location":"chapter_data_structure/classification_of_data_structure/","title":"Classification Of Data Structures","text":"

Common data structures include arrays, linked lists, stacks, queues, hash tables, trees, heaps, and graphs. They can be divided into two categories: logical structure and physical structure.

"},{"location":"chapter_data_structure/classification_of_data_structure/#logical-structures-linear-and-non-linear","title":"Logical Structures: Linear And Non-linear","text":"

Logical structures reveal logical relationships between data elements. In arrays and linked lists, data are arranged in sequential order, reflecting the linear relationship between data; while in trees, data are arranged hierarchically from the top down, showing the derived relationship between ancestors and descendants; and graphs are composed of nodes and edges, reflecting the complex network relationship.

As shown in the figure below, logical structures can further be divided into \"linear data structure\" and \"non-linear data structure\". Linear data structures are more intuitive, meaning that the data are arranged linearly in terms of logical relationships; non-linear data structures, on the other hand, are arranged non-linearly.

Non-linear data structures can be further divided into tree and graph structures.

"},{"location":"chapter_data_structure/classification_of_data_structure/#physical-structure-continuous-vs-dispersed","title":"Physical Structure: Continuous vs. Dispersed","text":"

When an algorithm is running, the data being processed is stored in memory. The figure below shows a computer memory module where each black square represents a memory space. We can think of the memory as a giant Excel sheet in which each cell can store data of a certain size.

The system accesses the data at the target location by means of a memory address. As shown in the figure below, the computer assigns a unique identifier to each cell in the table according to specific rules, ensuring that each memory space has a unique memory address. With these addresses, the program can access the data in memory.

Tip

It is worth noting that comparing memory to the Excel sheet is a simplified analogy. The actual memory working mechanism is more complicated, involving the concepts of address, space, memory management, cache mechanism, virtual and physical memory.

Memory is a shared resource for all programs, and when a block of memory is occupied by one program, it cannot be used by other programs at the same time. Therefore, considering memory resources is crucial in designing data structures and algorithms. For example, the algorithm's peak memory usage should not exceed the remaining free memory of the system; if there is a lack of contiguous memory blocks, then the data structure chosen must be able to be stored in non-contiguous memory blocks.

As shown in the figure below, Physical structure reflects the way data is stored in computer memory and it can be divided into consecutive space storage (arrays) and distributed space storage (linked lists). The physical structure determines how data is accessed, updated, added, deleted, etc. Logical and physical structure complement each other in terms of time efficiency and space efficiency.

It is worth stating that all data structures are implemented based on arrays, linked lists, or a combination of the two. For example, stacks and queues can be implemented using both arrays and linked lists; and implementations of hash tables may contain both arrays and linked lists.

Data structures based on arrays are also known as \"static data structures\", which means that such structures' length remains constant after initialization. In contrast, data structures based on linked lists are called \"dynamic data structures\", meaning that their length can be adjusted during program execution after initialization.

Tip

If you find it difficult to understand the physical structure, it is recommended that you read the next chapter, \"Arrays and Linked Lists,\" before reviewing this section.

"},{"location":"chapter_introduction/","title":"Chapter 1. \u00a0 Introduction to Algorithms","text":"

Abstract

A graceful maiden dances, intertwined with the data, her skirt swaying to the melody of algorithms.

She invites you to a dance, follow her steps, and enter the world of algorithms full of logic and beauty.

"},{"location":"chapter_introduction/#_1","title":"\u672c\u7ae0\u5185\u5bb9","text":""},{"location":"chapter_introduction/algorithms_are_everywhere/","title":"1.1 \u00a0 Algorithms are Everywhere","text":"

When we hear the word \"algorithm,\" we naturally think of mathematics. However, many algorithms do not involve complex mathematics but rely more on basic logic, which can be seen everywhere in our daily lives.

Before formally discussing algorithms, there's an interesting fact worth sharing: you have already unconsciously learned many algorithms and have become accustomed to applying them in your daily life. Here, I will give a few specific examples to prove this point.

Example 1: Looking Up a Dictionary. In an English dictionary, words are listed alphabetically. Suppose we're searching for a word that starts with the letter \\(r\\). This is typically done in the following way:

  1. Open the dictionary to about halfway and check the first letter on the page, let's say the letter is \\(m\\).
  2. Since \\(r\\) comes after \\(m\\) in the alphabet, we can ignore the first half of the dictionary and focus on the latter half.
  3. Repeat steps 1. and 2. until you find the page where the word starts with \\(r\\).
<1><2><3><4><5>

Figure 1-1 \u00a0 Process of Looking Up a Dictionary

This essential skill for elementary students, looking up a dictionary, is actually the famous \"Binary Search\" algorithm. From a data structure perspective, we can consider the dictionary as a sorted \"array\"; from an algorithmic perspective, the series of actions taken to look up a word in the dictionary can be viewed as \"Binary Search.\"

Example 2: Organizing Playing Cards. When playing cards, we need to arrange the cards in our hand in ascending order, as shown in the following process.

  1. Divide the playing cards into \"ordered\" and \"unordered\" sections, assuming initially the leftmost card is already in order.
  2. Take out a card from the unordered section and insert it into the correct position in the ordered section; after this, the leftmost two cards are in order.
  3. Continue to repeat step 2. until all cards are in order.

Figure 1-2 \u00a0 Playing Cards Sorting Process

The above method of organizing playing cards is essentially the \"Insertion Sort\" algorithm, which is very efficient for small datasets. Many programming languages' sorting functions include the insertion sort.

Example 3: Making Change. Suppose we buy goods worth \\(69\\) yuan at a supermarket and give the cashier \\(100\\) yuan, then the cashier needs to give us \\(31\\) yuan in change. They would naturally complete the thought process as shown below.

  1. The options are currencies smaller than \\(31\\), including \\(1\\), \\(5\\), \\(10\\), and \\(20\\).
  2. Take out the largest \\(20\\) from the options, leaving \\(31 - 20 = 11\\).
  3. Take out the largest \\(10\\) from the remaining options, leaving \\(11 - 10 = 1\\).
  4. Take out the largest \\(1\\) from the remaining options, leaving \\(1 - 1 = 0\\).
  5. Complete the change-making, with the solution being \\(20 + 10 + 1 = 31\\).

Figure 1-3 \u00a0 Change making process

In the above steps, we make the best choice at each step (using the largest denomination possible), ultimately resulting in a feasible change-making plan. From the perspective of data structures and algorithms, this method is essentially a \"Greedy\" algorithm.

From cooking a meal to interstellar travel, almost all problem-solving involves algorithms. The advent of computers allows us to store data structures in memory and write code to call the CPU and GPU to execute algorithms. In this way, we can transfer real-life problems to computers, solving various complex issues more efficiently.

Tip

If concepts such as data structures, algorithms, arrays, and binary search still seem somewhat obsecure, I encourage you to continue reading. This book will gently guide you into the realm of understanding data structures and algorithms.

"},{"location":"chapter_introduction/summary/","title":"1.3 \u00a0 Summary","text":""},{"location":"chapter_introduction/what_is_dsa/","title":"1.2 \u00a0 What is an Algorithm","text":""},{"location":"chapter_introduction/what_is_dsa/#121-definition-of-an-algorithm","title":"1.2.1 \u00a0 Definition of an Algorithm","text":"

An \"algorithm\" is a set of instructions or steps to solve a specific problem within a finite amount of time. It has the following characteristics:

"},{"location":"chapter_introduction/what_is_dsa/#122-definition-of-a-data-structure","title":"1.2.2 \u00a0 Definition of a Data Structure","text":"

A \"data structure\" is a way of organizing and storing data in a computer, with the following design goals:

Designing data structures is a balancing act, often requiring trade-offs. If you want to improve in one aspect, you often need to compromise in another. Here are two examples:

"},{"location":"chapter_introduction/what_is_dsa/#123-relationship-between-data-structures-and-algorithms","title":"1.2.3 \u00a0 Relationship Between Data Structures and Algorithms","text":"

As shown in the Figure 1-4 , data structures and algorithms are highly related and closely integrated, specifically in the following three aspects:

Figure 1-4 \u00a0 Relationship between data structures and algorithms

Data structures and algorithms can be likened to a set of building blocks, as illustrated in the Figure 1-5 . A building block set includes numerous pieces, accompanied by detailed assembly instructions. Following these instructions step by step allows us to construct an intricate block model.

Figure 1-5 \u00a0 Assembling blocks

The detailed correspondence between the two is shown in the Table 1-1 .

Table 1-1 \u00a0 Comparing Data Structures and Algorithms to Building Blocks

Data Structures and Algorithms Building Blocks Input data Unassembled blocks Data structure Organization of blocks, including shape, size, connections, etc Algorithm A series of steps to assemble the blocks into the desired shape Output data Completed Block model

It's worth noting that data structures and algorithms are independent of programming languages. For this reason, this book is able to provide implementations in multiple programming languages.

Conventional Abbreviation

In real-life discussions, we often refer to \"Data Structures and Algorithms\" simply as \"Algorithms\". For example, the well-known LeetCode algorithm problems actually test both data structure and algorithm knowledge.

"},{"location":"chapter_preface/","title":"Chapter 0. \u00a0 Preface","text":"

Abstract

Algorithms are like a beautiful symphony, with each line of code flowing like a rhythm.

May this book ring softly in your mind, leaving a unique and profound melody.

"},{"location":"chapter_preface/#_1","title":"\u672c\u7ae0\u5185\u5bb9","text":""},{"location":"chapter_preface/about_the_book/","title":"0.1 \u00a0 About This Book","text":"

This open-source project aims to create a free, and beginner-friendly crash course on data structures and algorithms.

"},{"location":"chapter_preface/about_the_book/#011-target-audience","title":"0.1.1 \u00a0 Target Audience","text":"

If you are new to algorithms with limited exposure, or you have accumulated some experience in algorithms, but you only have a vague understanding of data structures and algorithms, and you are constantly jumping between \"yep\" and \"hmm\", then this book is for you!

If you have already accumulated a certain amount of problem-solving experience, and are familiar with most types of problems, then this book can help you review and organize your algorithm knowledge system. The repository's source code can be used as a \"problem-solving toolkit\" or an \"algorithm cheat sheet\".

If you are an algorithm expert, we look forward to receiving your valuable suggestions, or join us and collaborate.

Prerequisites

You should know how to write and read simple code in at least one programming language.

"},{"location":"chapter_preface/about_the_book/#012-content-structure","title":"0.1.2 \u00a0 Content Structure","text":"

The main content of the book is shown in the following figure.

Figure 0-1 \u00a0 Main Content of the Book

"},{"location":"chapter_preface/about_the_book/#013-acknowledgements","title":"0.1.3 \u00a0 Acknowledgements","text":"

Throughout the creation of this book, numerous individuals provided invaluable assistance, including but not limited to:

Throughout the writing journey, I delved into numerous textbooks and articles on data structures and algorithms. These works served as exemplary models, ensuring the accuracy and quality of this book's content. I extend my gratitude to all who preceded me for their invaluable contributions!

This book advocates a combination of hands-on and minds-on learning, inspired in this regard by \"Dive into Deep Learning\". I highly recommend this excellent book to all readers.

Heartfelt thanks to my parents, whose ongoing support and encouragement have allowed me to do this interesting work.

"},{"location":"chapter_preface/suggestions/","title":"0.2 \u00a0 How To Read","text":"

Tip

For the best reading experience, it is recommended that you read through this section.

"},{"location":"chapter_preface/suggestions/#021-conventions-of-style","title":"0.2.1 \u00a0 Conventions Of Style","text":" PythonC++JavaC#GoSwiftJSTSDartRustCZig 
\"\"\"Header comments for labeling functions, classes, test samples, etc.\"\"\"\"\n\n# Content comments for detailed code solutions\n\n\"\"\"\nmulti-line\nmarginal notes\n\"\"\"\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
/* Header comments for labeling functions, classes, test samples, etc. */\n\n// Content comments for detailed code solutions.\n\n/**\n * multi-line\n * marginal notes\n */\n
// Header comments for labeling functions, classes, test samples, etc.\n\n// Content comments for detailed code solutions.\n\n// Multi-line\n// Annotation\n
"},{"location":"chapter_preface/suggestions/#022-learn-efficiently-in-animated-graphic-solutions","title":"0.2.2 \u00a0 Learn Efficiently In Animated Graphic Solutions","text":"

Compared with text, videos and pictures have a higher degree of information density and structure and are easier to understand. In this book, key and difficult knowledge will be presented mainly in the form of animations and graphs, while the text serves as an explanation and supplement to the animations and graphs.

If, while reading the book, you find that a particular paragraph provides an animation or a graphic solution as shown below, please use the figure as the primary source and the text as a supplement and synthesize the two to understand the content.

Figure 0-2 \u00a0 Example animation

"},{"location":"chapter_preface/suggestions/#023-deeper-understanding-in-code-practice","title":"0.2.3 \u00a0 Deeper Understanding In Code Practice","text":"

The companion code for this book is hosted in the GitHub repository. As shown in the Figure 0-3 , the source code is accompanied by test samples that can be run with a single click.

If time permits, it is recommended that you refer to the code and knock it through on your own. If you have limited time to study, please read through and run all the code at least once.

The process of writing code is often more rewarding than reading it. Learning by doing is really learning.

Figure 0-3 \u00a0 Running code example

The preliminaries for running the code are divided into three main steps.

Step 1: Install the local programming environment. Please refer to Appendix Tutorial for installation, or skip this step if already installed.

Step 2: Clone or download the code repository. If Git is already installed, you can clone this repository with the following command.

git clone https://github.com/krahets/hello-algo.git\n

Of course, you can also in the location shown in the Figure 0-4 , click \"Download ZIP\" directly download the code zip, and then in the local solution.

Figure 0-4 \u00a0 Clone repository with download code

Step 3: Run the source code. As shown in the Figure 0-5 , for the code block labeled with the file name at the top, we can find the corresponding source code file in the codes folder of the repository. The source code files can be run with a single click, which will help you save unnecessary debugging time and allow you to focus on what you are learning.

Figure 0-5 \u00a0 Code block with corresponding source file

"},{"location":"chapter_preface/suggestions/#024-growing-together-in-questioning-and-discussion","title":"0.2.4 \u00a0 Growing Together In Questioning And Discussion","text":"

While reading this book, please don't skip over the points that you didn't learn. Feel free to ask your questions in the comment section. We will be happy to answer them and can usually respond within two days.

As you can see in the Figure 0-6 , each post comes with a comment section at the bottom. I hope you'll pay more attention to the comments section. On the one hand, you can learn about the problems that people encounter, so as to check the gaps and stimulate deeper thinking. On the other hand, we expect you to generously answer other partners' questions, share your insights, and help others improve.

Figure 0-6 \u00a0 Example of comment section

"},{"location":"chapter_preface/suggestions/#025-algorithm-learning-route","title":"0.2.5 \u00a0 Algorithm Learning Route","text":"

From a general point of view, we can divide the process of learning data structures and algorithms into three stages.

  1. Introduction to Algorithms. We need to familiarize ourselves with the characteristics and usage of various data structures and learn about the principles, processes, uses and efficiency of different algorithms.
  2. Brush up on algorithm questions. It is recommended to start brushing from popular topics, such as Sword to Offer and LeetCode Hot 100, first accumulate at least 100 questions to familiarize yourself with mainstream algorithmic problems. Forgetfulness can be a challenge when first brushing up, but rest assured that this is normal. We can follow the \"Ebbinghaus Forgetting Curve\" to review the questions, and usually after 3-5 rounds of repetitions, we will be able to memorize them.
  3. Build the knowledge system. In terms of learning, we can read algorithm column articles, solution frameworks and algorithm textbooks to continuously enrich the knowledge system. In terms of brushing, we can try to adopt advanced brushing strategies, such as categorizing by topic, multiple solutions, multiple solutions, etc. Related brushing tips can be found in various communities.

As shown in the Figure 0-7 , this book mainly covers \"Phase 1\" and is designed to help you start Phase 2 and 3 more efficiently.

Figure 0-7 \u00a0 algorithm learning route

"},{"location":"chapter_preface/summary/","title":"0.3 \u00a0 Summary","text":""}]}